M.C.Q (1 Marks)

MCQ 11 Mark

The principal value of $\tan^{-1}\Big(\tan\frac{3\pi}{5}\Big)$ is:

A
$\frac{2\pi}{5}$
✓
$\frac{-2\pi}{5}$
C
$\frac{3\pi}{5}$
D
$\frac{-3\pi}{5}$

Answer

Correct option: B.

$\frac{-2\pi}{5}$

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MCQ 21 Mark

$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:

A
$\frac{6}{25}$
B
$\frac{24}{25}$
C
$\frac{4}{5}$
✓
$-\frac{24}{25}$

Answer

Correct option: D.

$-\frac{24}{25}$

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MCQ 31 Mark

$\tan^{-1}(\sqrt{3})$

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{3}$
C
$\frac{2\pi}{3}$
D
$\frac{5\pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{3}$

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MCQ 41 Mark

If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:

✓
$\frac{1}{\sqrt2}<\text{x}\leq1$
B
$0\leq\text{x}\leq\frac{1}{\sqrt2}$
C
$-1\leq\text{x}<\frac{1}{\sqrt2}$
D
$\text{x}>0$

Answer

Correct option: A.

$\frac{1}{\sqrt2}<\text{x}\leq1$

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MCQ 51 Mark

Choose the correct answer from the given four options.The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$ is:

A
$2+\sqrt{5}$
✓
$\sqrt{5}-2$
C
$\frac{\sqrt{5}+2}{2}$
D
$5+\sqrt{2}$ Hint: $\bigg[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\bigg]$

Answer

Correct option: B.

$\sqrt{5}-2$

We have, $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}\Big)$
Let $\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}=\theta$
$\Rightarrow\ \cos^{-1}\frac{2}{\sqrt{5}}=2\theta$
$\Rightarrow\ \cos2\theta=\frac{2}{\sqrt{5}}$
$\therefore\ 2\cos^{2}\theta-1=\frac{2}{\sqrt{5}}$
$\Rightarrow\ \cos^2\theta=\frac{1}{2}+\frac{1}{\sqrt{5}}$
$\Rightarrow\ \cos\theta=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}$
$\therefore\ \tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\sqrt{\frac{\frac{1}{2}-\frac{1}{\sqrt{5}}}{\frac{1}{2}+\frac{1}{\sqrt{5}}}}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$
$=\sqrt{\frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)}}=\sqrt{5}-2$

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MCQ 61 Mark

If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of $x$ is:

A
$0$
B
$-2$
C
$1$
✓
$2$

Answer

Correct option: D.

$2$

$\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}+1}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}}{1-\frac{\text{x}+1}{\text{x}-1}\times\frac{\text{x}-1}{\text{x}}}\Bigg)=\tan^{-1}(-7)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}^2+\text{x}+\text{x}^2-2\text{x}+1}{\text{x}^2-\text{x}-(\text{x}^2-1)}\Big)\tan^{-1}(-7)$
$\Rightarrow\frac{2\text{x}^2-\text{x}+1}{-\text{x}+1}=-7$
$\Rightarrow2\text{x}^2-\text{x}+1=7\text{x}-7$
$\Rightarrow2\text{x}^2-8\text{x}+8=0$
$\Rightarrow\text{x}^2-4\text{x}+4=0$
$\Rightarrow(\text{x}-2)^2=0$
$\Rightarrow\text{x}=2$

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MCQ 71 Mark

$\sin^{-1}\Big(\frac{-1}{2}\Big)$

A
$\frac{\pi}{3}$
B
$-\frac{\pi}{3}$
C
$\frac{\pi}{6}$
✓
$-\frac{\pi}{6}$

Answer

Correct option: D.

$-\frac{\pi}{6}$

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MCQ 81 Mark

The number of solutions of the equation $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$ is:

✓
$2$
B
$3$
C
$1$
D
None of these

Answer

Correct option: A.

$2$

We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big).$
$\therefore \tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}\Big)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}=1$
$\Rightarrow5\text{x}=1-6\text{x}^2$
$\Rightarrow6\text{x}^2+5\text{x}-1=0$

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MCQ 91 Mark

If $ \frac{3\text{x}+1}{(\text{x}-1)(\text{x}+3)} = \frac{\text{A}}{\text{x}-1}+\frac{B}{\text{x}+3} $ then $ {\sin}^{-1} \frac{\text{A}}{\text{B}} :$

A
$ \frac{\pi}{2}$
B
$ \frac{\pi}{3}$
✓
$ \frac{\pi}{6}$
D
$ \frac{\pi}{8}$

Answer

Correct option: C.

$ \frac{\pi}{6}$

We have,$ \frac{3\text{x}+1}{(\text{x}-1)(\text{x}+3)} = \frac{\text{A}}{\text{x}-1}+\frac{B}{\text{x}+3} $
$\Rightarrow 3x + 1 = A (x + 3) + B(x - 1)$
Substitute $x = 1$ both sides
$\Rightarrow 3(1) + 1 = A(1 + 3) + 0$
$ \Rightarrow A = 1$
Substitute $x = - 3x$ both sides
$\Rightarrow 3( -3) + 1 = 0 + B(-3 -1)$
$\Rightarrow -8 - 4B$
$ \Rightarrow B = 2$
Hence $ \sin^{-1}\frac{\text{A}}{\text{B}}=\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$

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MCQ 101 Mark

If $\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\text{x}\in(0,1),$ then the value of $x$ is:

A
$0$
B
$\frac{\text{a}}{2}$
C
$\text{a}$
✓
$\frac{2\text{a}}{1-\text{a}^2}$

Answer

Correct option: D.

$\frac{2\text{a}}{1-\text{a}^2}$

$\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Let, $\text{a}=\tan\theta\Rightarrow\theta=\tan^{-1}\text{a}$
$\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)=2\tan^{-1}(\text{x})$
$2\theta+2\theta=2\tan^{-1}(\text{x})$
$4\theta=2\tan^{-1}(\text{x})$
$2\tan^{-1}\text{a}=\tan^{-1}(\text{x})$
$\tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)=\tan^{-1}(\text{x})$
$\text{x}=\frac{2\text{a}}{1-\text{a}^2}$

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MCQ 111 Mark

Choose the correct answer from the given four options.The value of $\cos^{-1}\Big(\cos\frac{3\pi}{2}\Big)$ is equal to:

✓
$\frac{\pi}{2}$
B
$\frac{3\pi}{2}$
C
$\frac{5\pi}{2}$
D
$\frac{7\pi}{2}$

Answer

Correct option: A.

$\frac{\pi}{2}$

We have, $\cos^{-1}\Big(\cos\frac{3\pi}{2}\Big)$
$=\cos^{-1}\cos\Big(2\pi-\frac{\pi}{2}\Big)$
$=\cos^{-1}\cos\Big(\frac{\pi}{2}\Big)$
$[\because\ \cos(2\pi-\theta)=\cos\theta]$
$=\frac{\pi}{2}\ \Big[\because\ \cos^{-1}(\cos\text{x})=\text{x},\ \text{x}\in[0,\pi]\Big]$

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MCQ 121 Mark

$2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$ is equal to:

A
$\cot^{-1}\text{x}$
B
$\cot^{-1}\text{x}$
✓
$\tan^{-1}\text{x}$
D
None of these

Answer

Correct option: C.

$\tan^{-1}\text{x}$

$\therefore\ 2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$
$=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\Big(\tan^{-1}\frac{1}{\text{x}}\Big)\Big\}$
$=\frac{3}{29}=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\frac{1}{\text{x}}\Big\}$
$=2\tan^{-1}\Big\{\text{cosec y}-\frac{1}{\tan\text{y}}\Big\}$
$=2\tan^{-1}\Big\{\frac{1-\cos\text{y}}{\sin\text{y}}\Big\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{\sin\text{y}}\bigg\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{2\sin\frac{\text{y}}{2}\cos\frac{\text{y}}{2}}\bigg\}$
$=2\tan^{-1}\Big\{\tan\frac{\text{y}}{2}\Big\}$
$=\text{y}$
$=\tan^{-1}\text{x}$

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MCQ 131 Mark

Choose the correct answer from the given four options.Which of the following is the principal value branch of $\operatorname{cosec}^{-1} x ?$

A
$\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
B
$[0,\pi]-\Big\{\frac{\pi}{2}\Big\}$
C
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
✓
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$

Answer

Correct option: D.

$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$

We know that, the principal value branch of $\operatorname{cosec}^{-1} x ?$ is $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]-\{0\}$

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MCQ 141 Mark

Choose the correct answer from the given four options.If $\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\ \text{x}\in[0,1]$ then the value of $x$ is:

A
$0$
B
$\frac{\text{a}}{2}$
C
$\text{a}$
✓
$\frac{2\text{a}}{1-\text{a}^2}$

Answer

Correct option: D.

$\frac{2\text{a}}{1-\text{a}^2}$

We have, $\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$\Rightarrow\ 2\tan^{-1}\text{a}+2\tan^{-1}\text{a}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$\bigg[\because\ 2\tan^{-1}\text{a}=\sin^{-1}\Big(\frac{2\text{a}}{1+\text{a}^2}\Big)=\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)\bigg]$
$\Rightarrow\ 4\tan^{-1}\text{a}=2\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\ 2\tan^{-1}\text{a}=\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\ \tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)=\tan^{-1}\text{x}$
$\Big[\because\ 2\tan^{-1}\text{a}=\tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)\Big]$
$\Rightarrow\ \text{x}=\frac{2\text{a}}{1-\text{a}^2}$

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MCQ 151 Mark

If $\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=\text{x},$ then $\sin\text{x}$ is equal to:

✓
$\tan^2\Big(\frac{\alpha}{2}\Big)$
B
$\cot^2\Big(\frac{\alpha}{2}\Big)$
C
$\tan\alpha$
D
$\cot\Big(\frac{\alpha}{2}\Big)$

Answer

Correct option: A.

$\tan^2\Big(\frac{\alpha}{2}\Big)$

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MCQ 161 Mark

If ${ \sin }^{ -1 }\frac { \text{x} }{ 5 } +{ \text{cosec} }^{ -1 }\frac { 5 }{ 4 } $ then x is equal to:

A
$1$
B
$4$
✓
$3$
D
$5$

Answer

Correct option: C.

$3$

${ \sin }^{ -1 }\frac { x }{ 5 } +{ \text{cosec} }^{ -1 }\frac { 5 }{ 4 } =\frac{ \pi }{ 2 }$
$ \Rightarrow { \sin }^{ -1 }\frac { \text{x} }{ 5 } +{ \sin }^{ -1 }\frac { 4 }{ 5 } =\frac { \pi }{ 2 }$
$ \Rightarrow \sin^{-1}\frac{\text{x}}{5}=\frac{\pi}{2}-\sin^{-1}\frac{4}{5}=\cos^{-1}\frac{4}{5}$
$ \Rightarrow \text{x}=5\sin\cos^{-1}\frac{4}{5}=5\sin\sin^{-1}\frac{3}{5}=3$

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MCQ 171 Mark

What is $ \tan ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +\tan ^{ -1 }{ \left( \frac { 1 }{ 3 } \right) }$equal to?

A
$ \frac { \pi }{ 3 }$
✓
$ \frac { \pi }{ 4 }$
C
$ \frac { \pi }{ 6 }$
D
$ \frac { \pi }{ 9 }$

Answer

Correct option: B.

$ \frac { \pi }{ 4 }$

We know the formula $ \tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\left(\frac { \text{a}+\text{b} }{ 1-\text{ab} } \right)$
So $\tan^{-1}\big(\frac{1}{2}\big)+\tan^{-1}\big(\frac{1}{3}\big)=\tan^{-1}\Bigg(\frac{\big(\frac{1}{2}\big)+\big(\frac{1}{3}\big)}{1-\big(\frac{1}{2}\big)\big(\frac{1}{2}\big)}\Bigg)$
$=\tan^{-1}\Bigg(\frac{\big(\frac{5}{6}\big)}{\big(\frac{5}{6}\big)}\Bigg)=\tan^{-1}(1)=\frac{\pi}{4}$

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MCQ 181 Mark

If $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4},$ then $x$ is:

✓
$\frac{1}{6}$
B
$1$
C
$(\frac{1}{6},-1)$
D
None of these.

Answer

Correct option: A.

$\frac{1}{6}$

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MCQ 191 Mark

The value of $\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$ is:

A
$\frac{33}{5}$
✓
$-\frac{\pi}{10}$
C
$\frac{\pi}{10}$
D
$\frac{7\pi}{5}$

Answer

Correct option: B.

$-\frac{\pi}{10}$

$\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$
$=\sin^{-1}\Big(\cos\Big(6\pi+\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\cos\Big(\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(\frac{\pi}{2}-\frac{3\pi}{5}\Big)\Big)$
$=\frac{\pi}{2}-\frac{3\pi}{5}$
$=\frac{-\pi}{10}$

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MCQ 201 Mark

If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$

✓
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$-\frac{\pi}{3}$

Answer

Correct option: A.

$\frac{\pi}{6}$

We have
$\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
Now, $\alpha-\beta=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-1}\Big)-\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}-\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}{1+{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\times\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}}\end{pmatrix}$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{\sqrt{3}\text{y}(2\text{y}-\text{x})+\sqrt{3}\text{x}(2\text{z}-\text{y})}}{\sqrt{3}\text{y}(2\text{y}-\text{x})}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{2\sqrt3\text{y}^2-\sqrt3\text{xy}+2\sqrt3\text{x}^2-\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{2\text{y}^2+2\text{x}^2-2\text{xy}}{2\sqrt3\text{y}^2+2\sqrt3\text{x}^2-2\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=\frac{\pi}{6}$

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MCQ 211 Mark

If $\text{x}=\sin ^{ -1 }{ \text{K} },\text{y}=\cos ^{ -1 }\text{K}, -1\le \text{K}\le 1$, then the correct relationship is:

A
$\text{x}+\text{y}=\frac{\pi}{8}$
B
$\text{x}+\text{y}={2}$
✓
$\text{x}+\text{y}=\frac{\pi}{2}$
D
$\text{x}+\text{y}=\frac{\pi}{8}$

Answer

Correct option: C.

$\text{x}+\text{y}=\frac{\pi}{2}$

$\because \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$
$\therefore \text{x}+\text{y}=\sin ^{ -1 }{ \text{K} } +\cos ^{ -1 }{ \text{K} } =\frac { \pi }{ 2 }$

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MCQ 221 Mark

If $\sin^{-1}(\text{x}^2-7\text{x}+12)=\text{n}\pi,\forall\text{ n }\in\text{ I},$ then $x =$

A
$-2$
✓
$4$
C
$-3$
D
$5$

Answer

Correct option: B.

$4$

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MCQ 231 Mark

Choose the correct answer from the given four options.The value of $\sin\big[2\tan^{-1}(0.75)\big]$ is equal to:

A
$0.75$
B
$1.5$
✓
$0.96$
D
$\sin1.5$

Answer

Correct option: C.

$0.96$

We have, $\sin\big[2\tan^{-1}(0.75)\big]$
$=\sin\Big(2\tan^{-1}\frac{3}{4}\Big)$
$=\sin\Bigg(\sin^{-1}\frac{2.\frac{3}{4}}{1+\frac{9}{16}}\Bigg)$
$\Big(\because\ 2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\sin\Bigg(\sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}}\Bigg)$
$=\sin\Big(\sin^{-1}\frac{24}{25}\Big)=\frac{24}{25}=0.96$

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MCQ 241 Mark

Choose the correct answer from the given four options.The value of $\sin^{-1}\bigg[\cos\Big(\frac{33\pi}{5}\Big)\bigg]$ is:

A
$\frac{3\pi}{5}$
B
$\frac{-7\pi}{5}$
C
$\frac{\pi}{10}$
✓
$\frac{-\pi}{10}$

Answer

Correct option: D.

$\frac{-\pi}{10}$

We have, $\sin^{-1}\bigg[\cos\Big(\frac{33\pi}{5}\Big)\bigg]=\sin^{-1}\bigg[\cos\Big(6\pi+\frac{33\pi}{5}\Big)\bigg]$
$=\sin^{-1}\bigg[\cos\Big(\frac{3\pi}{5}\Big)\bigg]$
$\Big[\because\ \cos(2\text{n}\pi+\theta)=\cos\theta\Big]$
$=\sin^{-1}\Big[\cos\Big(\frac{\pi}{2}+\frac{\pi}{10}\Big)\Big]$
$=\sin^{-1}\Big(-\sin\frac{\pi}{10}\Big)$
$=-\sin^{-1}\Big(\sin\frac{\pi}{10}\Big)$
$[\because\ \sin^{-1}(-\text{x})=-\sin^{-1}\text{x}]$
$=-\frac{\pi}{10}\ \Big[\because\ \sin^{-1}(\sin\text{x})=\text{x},\ \text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)\Big]$

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MCQ 251 Mark

$\sin\begin{Bmatrix}2\cos^{-1}\Big(\frac{-3}{5}\Big)\end{Bmatrix}$ is equal to:

A
$\frac{6}{25}$
B
$\frac{24}{25}$
C
$\frac{4}{5}$
✓
$-\frac{24}{25}$

Answer

Correct option: D.

$-\frac{24}{25}$

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MCQ 261 Mark

The range of the function, $\text{f(x)}=(1+\sec^{-1}\text{x})(1+\cos^{-1}\text{x})$ is:

A
$(-\infty,\infty)$
B
$(-\infty,0]\cup[4.\infty)$
C
$\big\{0,(1+\pi^2)\big\}$
✓
$[1.(1+\pi)^2]$

Answer

Correct option: D.

$[1.(1+\pi)^2]$

$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$
Here the limiting component is $\cos−1(\text{x}),$ since the domain of $\cos−1(\text{x}),$ is $[−1, 1].$
Therefore,
$\text{f}(1)=(1+0)(1+0)$
$=1$
$\text{f}(−1)=(1+\pi(1+\pi)$
$=(1+\pi)^2 $
Hence range of $\text{f(x)}=[1,(1+\pi)^2]$

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MCQ 271 Mark

The value of the expression $\tan\Big(\frac{1}{2}\cos^{-1}\frac{2}{\sqrt{3}}\Big)$

A
$2+\sqrt{5}$
✓
$\sqrt{5}-2$
C
$\frac{\sqrt{5}+2}{2}$
D
$5+\sqrt{2}$

Answer

Correct option: B.

$\sqrt{5}-2$

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MCQ 281 Mark

$\cos\Big(2\tan^{-1}\frac{1}{7}\Big)-\sin\Big(4\sin^{-1}\frac{1}{3}\Big)=$

A
$1$
✓
$0$
C
$\frac{1}{2}$
D
$-\frac{1}{2}$

Answer

Correct option: B.

$0$

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MCQ 291 Mark

Choose the correct answer from the given four options.The domain of the function $\cos ^{-1}(2 x-1)$ is:

✓
$[0,1]$
B
$[-1,1]$
C
$(-1,1)$
D
$[0,\pi]$

Answer

Correct option: A.

$[0,1]$

We have,$\cos ^{-1}(2 x-1)$
Now, we know that the domain of $\cos ^{-1}(x)$ is $-1\leq\text{x}\leq1$
$\therefore\ -1\leq2\text{x}-1\leq1$
Adding $1$ to all terms, we get
$\Rightarrow\ 0\leq2\text{x}\leq2$
Dividing all terms by $2,$ we get
$\Rightarrow\ 0\leq\text{x}\leq1$
$\therefore\ \text{x}\in[0,1]$

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MCQ 301 Mark

If $\cos \left ( 2\sin^{-1}\text{x} \right )=\frac{1}{9}$, the value of $x$ which satify equation is $ \pm \frac{a}{b}$. Find the value of $a + b:$

A
$2$
B
$3$
C
$4$
✓
$5$

Answer

Correct option: D.

$5$

Given, $\cos \left ( 2\sin^{-1}\text{x} \right )=\frac{1}{9}$
Let, $\sin^{-1}\text{x}=\theta .$
Then, $\cos 2\theta =\frac{1}{9}$
$ 1-2\sin ^{2}\theta =\frac{1}{9}$
or $1-2\text{x}^{2}=\frac{1}{9}$
$\text{x}^{2}=\frac{4}{9}\text{x}$
$ \therefore \text{a}+\text{b}=2+3=5$

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MCQ 311 Mark

Choose the correct answer from the given four options.If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:

A
$\frac{1}{\sqrt{2}}<\text{x}\leq1$
B
$0\leq\text{x}<\frac{1}{\sqrt{2}}$
✓
$-1\leq\text{x}<\frac{1}{\sqrt{2}}$
D
$\text{x}>0$

Answer

Correct option: C.

$-1\leq\text{x}<\frac{1}{\sqrt{2}}$

We have, $\cos^{-1}\text{x}>\sin^{-1}\text{x}$
$\Rightarrow\ \frac{\pi}{2}-\sin^{-1}\text{x}>\sin^{-1}\text{x}$
$\Rightarrow\ \frac{\pi}{2}>2\sin^{-1}\text{x}$
$\Rightarrow\ \sin^{-1}\text{x}<\frac{\pi}{4}\ ....(\text{i})$
But $-\frac{\pi}{2}\leq\sin^{-1}\text{x}\leq\frac{\pi}{2}\ ....(\text{ii})$
From $(i)$ and $(ii), -\frac{\pi}{2}\leq\sin^{-1}\text{x}<\frac{\pi}{4}$
$\Rightarrow\ \sin\Big(-\frac{\pi}{2}\Big)\leq\text{x}<\sin\frac{\pi}{4}$
$\Rightarrow\ -1\leq\text{x}<\frac{1}{\sqrt{2}}$

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MCQ 321 Mark

If $\tan^{-1}\Big(\frac{\text{x}-1}{\text{x}+2}\Big)+\tan^{-1}\Big(\frac{\text{x}+1}{\text{x}+2}\Big)=\frac{\pi}{4},$ then $x$ is equal to:

A
$\frac{1}{\sqrt{2}}$
B
$-\frac{1}{\sqrt2}$
✓
$\pm\sqrt{\frac{5}{2}}$
D
$\pm\frac{1}{2}$

Answer

Correct option: C.

$\pm\sqrt{\frac{5}{2}}$

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MCQ 331 Mark

If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$

✓
$\sin2\alpha$
B
$\sin\alpha$
C
$\cos2\alpha$
D
$\cos\alpha$

Answer

Correct option: A.

$\sin2\alpha$

$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{1+\text{x}^2-2\sqrt{1-\text{x}^2}\sqrt{1+\text{x}^2}+1-\text{x}^2}{1+\text{x}^2-1+\text{x}^2}=\tan\alpha$
$\frac{1-\sqrt{1-\text{x}^4}}{\text{x}^2}=\tan\alpha$
$1-\sqrt{1-\text{x}^4}=\text{x}^2\tan\alpha$
$\big(1-\text{x}^2\tan\alpha\big)^2=1-\text{x}^4$
$1-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=1-\text{x}^4$
$\text{x}^4-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=0$
$\text{x}^2\big(\text{x}^2-2\tan\alpha+\text{x}^2\tan^2\alpha\big)=0$
$\text{x}^2=\frac{2\tan\alpha}{1+\tan^2\alpha}$
$\text{x}^2=\frac{2\tan\alpha}{\sec^2\alpha}$
$\text{x}^2=2\tan\alpha\cos^2\alpha$
$\text{x}^2=2\sin\alpha\cos\alpha=2\sin\alpha$

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MCQ 341 Mark

The value of $ \cos^{-1}\left (\cot \left (\dfrac {\pi}{2}\right )\right ) + \cos^{-1} \left (\sin \left (\dfrac {2\pi}{3}\right )\right )$ is:

✓
$ \dfrac {2\pi}{3}$
B
$2$
C
$3$
D
$\pi $

Answer

Correct option: A.

$ \dfrac {2\pi}{3}$

$ \cos^{-1}\left (\cot \dfrac {\pi}{2}\right ) + \cos^{-1} \left (\sin \dfrac {2\pi}{3}\right ) = \cos^{-1} (0) + \cos^{-1} \left (\dfrac {\sqrt {3}}{2}\right )$
$=\frac{\pi}{2}+\cos^{-1}\bigg(\cos\frac{\pi}{6}\bigg)$
$ = \frac {\pi}{2} + \frac {\pi}{6}$
$ = \frac {4\pi}{6}$
$ = \frac {2\pi}{3}$

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MCQ 351 Mark

If $x < 0, y < 0$ such that $xy = 1,$ then $\tan^{-1}\text{x}+\tan^{-1}\text{y}$ equals:

A
$\frac{\pi}{2}$
✓
$-\frac{\pi}{2}$
C
$-\pi$
D
None of these

Answer

Correct option: B.

$-\frac{\pi}{2}$

We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$x < 0, y < 0$ such that
$xy = 1$
Let $x = -a$ and $y = -b,$ where $a$ and $b$ both are positive.
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{-\text{a}-\text{a}}{1-1}\Big)$
$=\tan^{-1}(-\infty)$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{2}\Big)\Big\}$
$=-\frac{\pi}{2}$

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MCQ 361 Mark

Find the value of $:\ \sec^2 (\tan^{-1} 2) +\csc^2 (\cot^{-1} 3)$

A
$11$
✓
$15$
C
$17$
D
$21$

Answer

Correct option: B.

$15$

$ \sec^2 (\tan^{-1} 2) +\csc^2 (\cot^{-1} 3)$
$ =1+\tan^2 (\tan^{-1} 2) +1+\cot^2 (\cot^{-1} 3)$
$ =1+[\tan (\tan^{-1} 2)]^2 +1+[\cot (\cot^{-1} 3)]^2$
$ =1+2^2+1+3^2=15$

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MCQ 371 Mark

$\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)=$

✓
$\frac{6}{17}$
B
$\frac{3}{17}$
C
$\frac{4}{17}$
D
$\frac{5}{17}$

Answer

Correct option: A.

$\frac{6}{17}$

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MCQ 381 Mark

The value of $\sin(2\tan^{-1}(0.75))$ is equal to:

A
$0.75$
B
$1.5$
✓
$0.96$
D
$\sin1.5$

Answer

Correct option: C.

$0.96$

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MCQ 391 Mark

Domain of $ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +\text{cosec} ^{ -1 }{ \text{x}}$ is:

A
$[-1, 1]$
B
$R$
C
$(-\infty , -1] \cup [1, \infty )$
✓
$\{-1, 1\}$

Answer

Correct option: D.

$\{-1, 1\}$

$ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +\text{cosec} ^{ -1 }{ \text{x}}$
Domain of $\cot^{−1}\text{x}=(−\infty ,\infty )$
Domain of $\cos^{−1}\text{x}=(−1,1)$
Domain of $ \text{cosec}^{-1}\text{x} = (-\infty, -1]\cup [1, \infty)c$
These function are in addition.So,
we have to take the intersection of all domains.So,
answer is $\{-1, 1\}$
concept: $ \text{f}(\text{x}) = \text{f}_1(\text{x}) +\text{f}_2(\text{x}) + ...+\text{f}_\text{n}(\text{x})$
domain of $ \text{f}(\text{x})$
Domain of $\text{f}_1(\text{x}) \cap$
domain of $\text{f}_2(\text{x}) \cap$
domain of $\text{f}_\text{n}(\text{x})$

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MCQ 401 Mark

The value of $\cot^{-1}9+\text{cosec}^{-1}(\frac{\sqrt{41}}{4})$ is given by:

A
$0$
✓
$\frac{\pi}{4}$
C
$\tan^{-1}2$
D
$\frac{\pi}{2}$

Answer

Correct option: B.

$\frac{\pi}{4}$

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MCQ 411 Mark

$\cot(\frac{\pi}{4}-2\cot^{-1}3)=$

✓
$7$
B
$6$
C
$5$
D
None of these.

Answer

Correct option: A.

$7$

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MCQ 421 Mark

The value of $\tan\Big\{\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{4}{\sqrt{17}}\Big\}$ is:

A
$\frac{\sqrt{29}}{3}$
B
$\frac{29}{3}$
C
$\frac{\sqrt3}{29}$
✓
$\frac{3}{29}$

Answer

Correct option: D.

$\frac{3}{29}$

Let, $\cos^{-1}\frac{1}{5\sqrt2}=\text{y}$ and $\sin^{-1}\frac{4}{\sqrt{17}}=\text{z}$
$\therefore\ \cos\text{y}=\frac{1}{5\sqrt2}$
$\Rightarrow\sin\text{y}=\frac{7}{5\sqrt2}$
$\Rightarrow\tan\text{y}=7$
$\sin\text{z}=\frac{4}{\sqrt{17}}$
$\Rightarrow\cos\text{z}=\frac{1}{\sqrt{17}}$
$\Rightarrow\tan\text{z}=4$
$\therefore\ \tan\Big(\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{1}{\sqrt{17}}\Big)=\tan(\text{y}-\text{z})$
$=\frac{\tan\text{y}-\tan\text{z}}{1+\tan\text{y}\tan\text{z}}$
$=\frac{7-4}{1+7\times4}$
$=\frac{3}{29}$

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MCQ 431 Mark

What will be the value of $ \text{x} + \text{y} + \text{z }$ if $\cos-1 \text{x} + \cos-1 \text{y} + \cos-1 \text{z} = 3\pi ?$

A
$ \frac{-1}{3}$
B
$1$
C
$3$
✓
$-3$

Answer

Correct option: D.

$-3$

The equation is $ \cos-1 \text{x} +\cos-1 \text{y} + \cos-1 \text{z} = 3\pi $
This means $ \cos-1 \text{x} = \pi , \cos-1 \text{y} = \pi $ and $ \cos-1 \text{z} = \pi $
This will be only possible when it is in maxima.
As, $\cos-1 \text{x} = \pi $ so,$ \text{x} = \cos-1 \pi = -1$ similarly$, y = z = -1$
Therefore$, x + y + z = -1 -1 -1$
So$, x + y + z = -3.$

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MCQ 441 Mark

If $\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}+\tan^{-1}(\text{x}+1)=\tan^{-1}3\text{x},$ then the values of $x$ are:

A
$\pm\frac{1}{2}$
B
$0,\frac{1}{2}$
C
$0,-\frac{1}{2}$
✓
$0,\pm\frac{1}{2}$

Answer

Correct option: D.

$0,\pm\frac{1}{2}$

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MCQ 451 Mark

$\cos[\tan^{-1}\{\sin(\cot^{-1}\text{x})\}]$ is equal to:

A
$\sqrt{\frac{\text{x}^2+2}{\text{x}^3+3}}$
B
$\sqrt{\frac{\text{x}^2+2}{\text{x}^2+1}}$
✓
$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
D
None of these.

Answer

Correct option: C.

$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$

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MCQ 461 Mark

$[-1, 1]$ is the domain for which of the following inverse trigonometric functions?

✓
$\sin^{-1}\text{⁡x}$
B
$\cot^{-1}\text{⁡x}$
C
$\tan^{-1}\text{⁡x}$
D
$\sec^{-1}\text{⁡x}$

Answer

Correct option: A.

$\sin^{-1}\text{⁡x}$

$[-1, 1]$ is the domain for $\sin^{-1}\text{⁡x}$
The domain for $\cot^{-1}\text{⁡x}$ is $(-\infty , \infty ).$
The domain for $\tan^{-1}\text{⁡x}$ is $(-\infty , \infty ).$
The domain for $\sec^{-1}\text{⁡x}$ is $(-\infty , -1) \cup (1, \infty ).$

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MCQ 471 Mark

The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 12)$ is:

A
$0$
B
$ \pi $
✓
$ 8\pi - 24$
D
none of these

Answer

Correct option: C.

$ 8\pi - 24$

$12$ rad lies in $4^{th}$ quadrant
$ \frac{7\pi}{2}<12<4\pi$
Let $\theta$ be an acute angle such that
$ 12+\theta=4\pi$
$\therefore 12=4\pi −\theta$ or $\theta=4\pi-12\theta =4\pi −12$
$ \cos^{-1}(\cos12)-\sin^{-1}(\sin12)$
$ =\cos^{-1}(\cos(4\pi-\theta))-\sin^{-1}(\sin(4\pi-\theta))$
$ =\cos^{-1}(\cos\theta)-\sin^{-1}(-\sin\theta)$
$=\cos^{-1}(\cos\theta)-\sin^{-1}(\sin(-\theta))$
$ =\theta-(-\theta)$
$ =2\theta$
$ =2(4\pi −24)$
$ =8\pi −24$
$ \therefore \cos−1(\cos12)−\sin−1(\sin12)=8\pi −24$

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MCQ 481 Mark

If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then $x =$

A
$\frac{1}{2}$
✓
$\frac{\sqrt{3}}{2}$
C
$-\frac{1}{2}$
D
$-\frac{\sqrt{3}}{2}$

Answer

Correct option: B.

$\frac{\sqrt{3}}{2}$

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MCQ 491 Mark

If $\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2},$ then, $4\text{x}^2-12\text{xy}\cos^2\frac{\theta}{2}+9\text{y}^2=$

A
$36$
B
$36-36\cos\theta$
✓
$18-18\cos\theta$
D
$18+18\cos\theta$

Answer

Correct option: C.

$18-18\cos\theta$

$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
$\Rightarrow\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2}$
$\Rightarrow\cos^{-1}\Bigg(\frac{\text{x}}{3}\times\frac{\text{y}}{2}-\sqrt{1-\Big(\frac{\text{x}}{3}\Big)^2}\sqrt{1-\Big(\frac{\text{y}}{2}\Big)^2}\Bigg)=\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}}{6}-\sqrt{1-\Big(\frac{\text{x}^2}{9}\Big)}\sqrt{1-\Big(\frac{\text{y}^2}{4}\Big)}=\cos\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}-6\cos\frac{\theta}{2}}{6}=\frac{\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}}{6}$
$\Rightarrow\text{xy}-6\cos\frac{\theta}{2}=\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}$
Taking square on both sides,
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=\big(9-\text{x}^2\big)\big(4-\text{y}^2\big)$
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=36-9\text{y}^2-4\text{x}^2+\text{x}^2\text{y}^2$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\cos^2\frac{\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\frac{1+\cos\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\cos^2\frac{\theta}{2}=18-18\cos\theta$

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MCQ 501 Mark

The value of $\sin\bigg[\cos^{-1}\Big(\frac{7}{25}\Big)\bigg]$ is:

A
$\frac{25}{24}$
B
$\frac{25}{7}$
✓
$\frac{24}{25}$
D
$\frac{7}{24}$

Answer

Correct option: C.

$\frac{24}{25}$

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MCQ 511 Mark

Solve for $x : \{\text{x}\cos(\cot^{-1}\text{x})+\sin(\cot^{-1}\text{x})\}^2=\frac{51}{50}$

A
$\frac{1}{\sqrt{2}}$
✓
$\frac{1}{5\sqrt{2}}$
C
$2\sqrt{2}$
D
$5\sqrt{2}$

Answer

Correct option: B.

$\frac{1}{5\sqrt{2}}$

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MCQ 521 Mark

If $\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta,$ then $\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2$ is equal to:

A
$36$
B
$-36\sin^2\theta$
✓
$36\sin^2\theta$
D
$-36\cos^2\theta$

Answer

Correct option: C.

$36\sin^2\theta$

We know
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big[\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big]$
Now,
$\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta$
$\Rightarrow\cos^{-1}\Big[\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}\Big]=\theta$
$\Rightarrow\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}=\cos\theta$
$\Rightarrow\text{xy}-\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=6\cos\theta$
$\Rightarrow\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=\text{xy}-6\cos\theta$
$\Rightarrow(4-\text{x}^2)(9-\text{y}^2)=\text{x}^2\text{y}^2+36\cos^2\theta-12\text{xy}\cos\theta\ ($Squaring both the sides$)$
$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^{2}\theta-12\text{xy}\cos\theta$
$\Rightarrow36-4\text{y}^2-9\text{x}^2=36\cos^2\theta-12\text{xy}\cos\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36-36\cos^2\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36\sin^2\theta$

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MCQ 531 Mark

$\Bigg(\cot\Bigg(\sin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}+\cos^{-1}\frac{\sqrt{-12}}{4}+\sec^{\sqrt{2}}\Bigg)\Bigg)$ is:

✓
$0$
B
$ 2\pi $
C
$ 3\pi $
D
None of these

Answer

Correct option: A.

$0$

The above expression can be rewritten as
$\sin^{-1}(\cot(15^{0}+30^{0}+45^{0}))$
$ =\sin^{-1}(\cot(90^{0}))$
$ =\sin^{-1}(0)$
$ = 0$

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MCQ 541 Mark

Choose the correct answer from the given four options. If $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi,$ then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ equals:

A
$0$
B
$1$
✓
$6$
D
$12$

Answer

Correct option: C.

$6$

The domain of $\cos ^{-1} x$ is $[0,\pi]$
We are given that, $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi$
Which is possible only when $\alpha=\beta=\gamma=\cos\pi\ $ or $-1$
Now, $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$
$=-1(-1-1)-1(-1-1)-1(-1-1)$
$=2+2+2$
$=6$

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MCQ 551 Mark

$\cos^{-1}[\cos(2\cot^{-1}(\sqrt2-1))]= ...........$

A
$\sqrt2-1$
B
$1+\sqrt2$
C
$\frac{\pi}{4}$
✓
$\frac{3\pi}{4}$

Answer

Correct option: D.

$\frac{3\pi}{4}$

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MCQ 561 Mark

The equation $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)$ has:

✓
Nique solution.
B
No solution.
C
Infinitely many solution.
D
None of these.

Answer

Correct option: A.

Nique solution.

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MCQ 571 Mark

What is the value of $ \sin-1(\sin 6)?$

A
$-2\pi - 6$
B
$2\pi + 6$
✓
either $-2\pi + 6$ or $2\pi + 6$
D
$2\pi - 6$

Answer

Correct option: C.

either $-2\pi + 6$ or $2\pi + 6$

We know that $\sin(\text{x}) = \sin(2\text{A}^ * \pi + \text{x})$
where $A$ can be positive or negative integer.
If $A$ is $-1,$ then $ \sin(6) = \sin(-2\pi + 6);$
If $A$ is $1,$ then $ \sin(6) = \sin(2\pi + 6).$

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MCQ 581 Mark

The value of $\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)=$

✓
$\frac{19}{8}$
B
$\frac{8}{19}$
C
$\frac{19}{2}$
D
$\frac{3}{4}$

Answer

Correct option: A.

$\frac{19}{8}$

$\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\Bigg(\tan^{-1}\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Bigg(\tan^{-1}\frac{\frac{4}{5}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Big(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\bigg(\tan^{-1}\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\bigg)$
$=\frac{\frac{16+3}{12}}{\frac{2}{3}}$
$=\frac{19}{8}$
Hence, the correct answer is option $(a).$

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MCQ 591 Mark

$ \tan^{−1}\sqrt{3}+\sec−12–\cos−^{1}1$ is equal to $...........$

A
$0$
✓
$ \frac{2}{\pi^3}$
C
$ \frac{\pi}{3}$
D
$ \frac{\pi}{4}$

Answer

Correct option: B.

$ \frac{2}{\pi^3}$

$\tan^{-1}\sqrt{3}=\frac{\pi}{3},\sec^{-1}2,\cos^{-1}1=0$
$ \therefore \tan^{−1}\sqrt{13}+\sec^{−1}2−\cos^{−1}1$
$=\frac{\pi }{3}+\frac{\pi }{3}−0$
$ =\frac{2\pi }{3}$

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MCQ 601 Mark

Consider $ \text{x} = 4\tan^{-1}\left (\frac {1}{5}\right ), \text{y} = \tan^{-1} \left (\frac {1}{70}\right ) $ and $ \text{z} = \tan^{-1}\bigg (\frac {1}{99}\bigg)$ .What is $xx$ equal to?

A
$ \tan^{-1}\left (\frac {60}{119}\right )$
✓
$ \tan^{-1}\left (\frac {120}{119}\right )$
C
$ \tan^{-1}\left (\frac {90}{119}\right )$
D
$ \tan^{-1}\left (\frac {170}{119}\right )$

Answer

Correct option: B.

$ \tan^{-1}\left (\frac {120}{119}\right )$

$\text{x}=4\ { \tan }^{ -1 }\left(\dfrac { 1 }{ 5 } \right)\\$
$ =2\ { \tan }^{ -1 }\left(\frac { 1 }{ 5 } \right)+2\ { \tan }^{ -1 }\left(\frac { 1 }{ 5 } \right)\\$
$ =2\ { \tan }^{ -1 }\left(\frac { \frac { 1 }{ 5 } +\frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 5 } \times \frac { 1 }{ 5 } } \right)\\$
$ =2\ { \tan }^{ -1 }\left(\frac { 5 }{ 12 } \right)\\$
$ =2\ { \tan }^{ -1 }\left(\frac {120 }{ 119 } \right)\\$

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MCQ 611 Mark

$3\tan^{-1} a$ is equal to:

A
$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1+3\text{a}^2}\Big)$
B
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1+3\text{a}^2}\Big)$
C
$\tan^{-1}\Big(\frac{3\text{a}+\text{a}^3}{1-3\text{a}^2}\Big)$
✓
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$

Answer

Correct option: D.

$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$

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MCQ 621 Mark

Consider the following statements:

$\tan^{-1} 1+ \tan^{-1} (0.5) = \dfrac {\pi}2$
$\sin^{-1}{\cfrac{1}{3} }+ \cos^{-1}{\cfrac{1}{3}} =\cfrac{\pi}{2}$

Which of the above statements is/are correct ?

A
$1$ only
✓
$2$ only
C
Both $1$ and $2$
D
Neither $1$ nor $2$

Answer

Correct option: B.

$2$ only

We know that $\tan^{-1} { \text{x} } + \cot^{-1} { \text{x} } =\frac { \pi }{ 2 }$$ \text{x} \in R$ and $\sin^{-1}{\text{x}} + \cos^{-1}{\text{x}} =\frac{\pi}{2}$,
and $ \sin-1\frac{1}{3}+\cos-1{\frac{1}{3}} =\cfrac{\pi}{2}$
Hence, only second statement is correct.

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MCQ 631 Mark

The range of $\sin^{-1}\text{x}+\cos^{-1}\text{x}+\tan^{-1}\text{x}$ is:

A
$[0,\pi]$
✓
$[\frac{\pi}{4},\frac{3\pi}{4}]$
C
$(0,\pi)$
D
$[0,\frac{\pi}{2}]$

Answer

Correct option: B.

$[\frac{\pi}{4},\frac{3\pi}{4}]$

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MCQ 641 Mark

Choose the correct answer from the given four options. Which of the following is the principal value branch of $\cos ^{-1} x?$

A
$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
B
$(0,\pi)$
✓
$[0,\pi]$
D
$(0,\pi)-\Big\{\frac{\pi}{2}\Big\}$

Answer

Correct option: C.

$[0,\pi]$

The principal value branch of $\cos ^{-1} x$ is $[0,\pi].$

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MCQ 651 Mark

$\cos^{-1}\frac{1}{2}+2\sin^{-1}\frac{1}{2}$ is equal to:

A
$\frac{\pi}{4}$
B
$\frac{\pi}{6}$
C
$\frac{\pi}{3}$
✓
$\frac{2\pi}{3}$

Answer

Correct option: D.

$\frac{2\pi}{3}$

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MCQ 661 Mark

The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$ is

A
$0$
✓
$8\pi - 26$
C
$4\pi + 2$
D
None of these

Answer

Correct option: B.

$8\pi - 26$

The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$
$ =\cos^{−1}(\cos(4\pi −12))−\sin^{−1}(−\sin(4\pi −14))$
$ =4\pi − 12 − 14 + 4\pi $
$= 8\pi − 26$

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MCQ 671 Mark

The value of $ \cos \left( \sin^{-1} \left( \frac {2}{3} \right) \right)$ is equal to :

A
$ \frac {\sqrt4}{8}$
B
$ \frac {\sqrt4}{3}$
C
$ \frac {\sqrt5}{4}$
✓
$ \frac {\sqrt5}{3}$

Answer

Correct option: D.

$ \frac {\sqrt5}{3}$

The value of $ \cos \left( \sin^{-1} \left( \frac {2}{3} \right) \right)$ is
$=\cos\Bigg(\cos^{-1}\sqrt{1-\bigg(1-\frac{2}{3}\bigg)^2}\Bigg)$
$ = \cos \left( \cos^{-1}\left( \sqrt{1 - \frac{4}{9} } \right) \right)$
$ = \cos \left( \cos^{-1} \left( \frac {\sqrt5}{3} \right) \right)$
$ = \frac {\sqrt5}{3}$

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MCQ 681 Mark

$ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$ is equal to

✓
$ \frac{\pi}{2}$
B
$ \frac{\pi}{3}$
C
$ \frac{\pi}{4}$
D
$ \frac{\pi}{6}$

Answer

Correct option: A.

$ \frac{\pi}{2}$

Given, $ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$
$\Rightarrow \sin^{−1}\text{x}+\sin^{−1}\text{y}=\sin ^{ -1 } \Big(\text{x}\sqrt { 1-{ \text{y} }^{ 2 } } +\text{y}\sqrt { 1-\text{x}^{ 2 } }\Big)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { 1-\left(\frac { 4 }{ 5 } \right) } +\frac { 4 }{ 5 } \sqrt { 1-\left(\frac { 3 }{ 5 } \right)^{ 2 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { \frac { 25-16 }{ 25 } ) } +\frac { 4 }{ 5 } \sqrt { \frac { 25-9 }{ 25 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \times \frac { 3 }{ 5 } +\frac { 4 }{ 5 } \times \frac { 4 }{ 5 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 16 }{ 25 } +\frac { 9 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 25 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } (1)$
$ \Rightarrow \cfrac { \pi }{ 2 }$

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MCQ 691 Mark

In a $\triangle\text{ABC},$ if $C$ is a right angle, then $\tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{b}}\Big)=$

A
$\frac{\pi}{3}$
✓
$\frac{\pi}{4}$
C
$\frac{5\pi}{2}$
D
$\frac{\pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{4}$

We know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$
$\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$

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MCQ 701 Mark

The number of real solution of the equation $\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$ is:

A
$0$
B
$1$
✓
$2$
D
infinite

Answer

Correct option: C.

$2$

$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$
$\Rightarrow\sqrt{2\cos^2\text{x}}=\sqrt2(-\pi-\text{x})$
$\Rightarrow|\cos\text{x}|=\text{x}$
If $\cos\text{x}$ is positive then $\cos\text{x}=-\pi-\text{x}$
It does not satisfy any value in the interval $\Big(-\pi,-\frac{\pi}{2}\Big)$
For the interval $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
$\cos\text{x}=\text{x}$
It gives the value of $x$ in the $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
For the interval $\Big[-\frac{\pi}{2},\pi\Big]$
$-\cos\text{x}=\pi-\text{x}$
$\cos\text{x}=\text{x}-\pi$
It gives one value of $x$ in the interval $\Big[\frac{\pi}{2},\pi\Big].$
Two real solution in the interval $[-\pi,\pi]$

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MCQ 711 Mark

The value of expression $2\sec^{-1}0+\sin^{-1}\Big(\frac{1}{2}\Big)$

A
$\frac{\pi}{6}$
✓
$\frac{5\pi}{6}$
C
$\frac{7\pi}{6}$
D
$1$

Answer

Correct option: B.

$\frac{5\pi}{6}$

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MCQ 721 Mark

Solve for $x : \sin^{-1}2\text{x}+\sin^{-1}3\text{x}=\frac{\pi}{3}$

A
$\sqrt{\frac{76}{3}}$
✓
$\sqrt{\frac{3}{76}}$
C
$\frac{3}{\sqrt{76}}$
D
$\frac{\sqrt{3}}{76}$

Answer

Correct option: B.

$\sqrt{\frac{3}{76}}$

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MCQ 731 Mark

$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$ is equal to:

A
$0$
B
$\frac{1}{2}$
C
$-1$
✓
None of these

Answer

Correct option: D.

None of these

$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{2}{11}\times\frac{1}{11}}\Bigg)$
$=\tan^{-1}\Bigg(\frac{\frac{3}{11}}{1-\frac{2}{121}}\Bigg)$
$=\tan^{-1}\Big(\frac{33}{119}\Big)$

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MCQ 741 Mark

$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$ is equal to:

✓
$\text{x}$
B
$\sqrt{1-\text{x}^2}$
C
$\frac{1}{\text{x}}$
D
None of these

Answer

Correct option: A.

$\text{x}$

Put $\cos^{-1}\text{x}=\text{u}$
$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$
$=\sin\big[\cot^{-1}\{\tan(\text{u})\}\big]$
$=\sin\Big[\cot^{-1}\Big\{\cot\Big(\frac{\pi}{2}-\text{u}\Big )\Big\}\Big]$
$=\sin\Big[\frac{\pi}{2}-\text{u}\Big]$
$=\cos\text{u}$
$=\text{x} \big(\therefore \cos^{-1}\text{x}=\text{u}\Rightarrow\text{x}=\cos\text{u}\big)$

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MCQ 751 Mark

If $ \text{x} \in \left ( \frac{3\pi}{2}, 2\pi \right )$ then the value of the expression $ \sin^{-1}[\cos({\cos^{-1}(\cos \, \text{x})}+\sin^{-1}(\sin \, \text{x}))]$ is:

A
$5$
✓
$ \frac{\pi}{2}$
C
$0$
D
$\pi$

Answer

Correct option: B.

$ \frac{\pi}{2}$

$\text{x}\in \left ( \frac{3\pi}{2}, 2\pi \right )$
Now, $ \cos^{-1}(\cos \,\text{ x})=2\pi −\text{x}$
and $ \sin^{-1}(\sin \, \text{x})=\text{x}-2\pi$
$ \therefore \cos^{−1}(\cos\text{x})+\sin−1(\sin\text{x})=0$
$\sin−1[\cos{\cos−1(\cos\text{x})+\sin−1(\sin\text{x})}]$
$ =\sin^{−1}{\cos(0)}$
$=\sin^{−1}(1)$
$=\frac{\pi }{2}$

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MCQ 761 Mark

Number of triplets $(x, y, z)$ satisfying $ \sin ^{-1}\text{x}+\sin ^{-1}\text{y}+\cos ^{-1}\text{z}=2\pi$ is:

✓
$1$
B
$0$
C
$2$
D
$\infty $

Answer

Correct option: A.

$1$

Let $f(x, y, z) =\sin ^{-1}\text{x}+\sin ^{-1}\text{y}+\cos^{-1}\text{z}$
It will attain the value $2\pi$ only if $ \sin ^{-1}\text{x}=\sin ^{-1}\text{y}=\frac{\pi }{2}$ and $\cos^{-1}\text{z}=\pi$
This ispossible only if $x = y = 1$ and $z = -1$
Hence there is only one solutionf $ (1, 1, -1)= 2\pi .$

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MCQ 771 Mark

If $\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of $x$ is:

A
$\frac{3}{2}$
B
$\frac{1}{\sqrt{2}}$
✓
$\frac{\sqrt{3}}{2}$
D
$\frac{2}{\sqrt{3}}$

Answer

Correct option: C.

$\frac{\sqrt{3}}{2}$

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MCQ 781 Mark

The value of $\cot\Big(\text{cosec}^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\Big)$ is:

A
$\frac{5}{17}$
✓
$\frac{6}{17}$
C
$\frac{3}{17}$
D
$\frac{4}{17}$

Answer

Correct option: B.

$\frac{6}{17}$

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MCQ 791 Mark

$2\cos^{-1}\text{x}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$ is true for:

A
all $x$
B
$x > 0$
C
$\text{x }\in[-1,1]$
✓
$\frac{1}{\sqrt{2}}\leq\text{x}\leq1$

Answer

Correct option: D.

$\frac{1}{\sqrt{2}}\leq\text{x}\leq1$

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MCQ 801 Mark

$2\tan^{-1}(\cos\text{x})=\tan^{-1}(2\ \text{cosec x})$

A
$0$
B
$\frac{\pi}{3}$
✓
$\frac{\pi}{4}$
D
$\frac{\pi}{2}$

Answer

Correct option: C.

$\frac{\pi}{4}$

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MCQ 811 Mark

If $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8,$ then $x =$

A
$5$
✓
$\frac{1}{5}$
C
$\frac{5}{14}$
D
$\frac{14}{5}$

Answer

Correct option: B.

$\frac{1}{5}$

We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}$
Now, $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8$
$\Rightarrow\tan^{-1}\Big(\frac{3+\text{x}}{1-3\text{x}}\Big)=\tan^{-1}8$
$\Rightarrow\frac{3+\text{x}}{1-3\text{x}}=8$
$\Rightarrow3+\text{x}=8-24\text{x}$
$\Rightarrow3-8=-24\text{x}-\text{x}$
$\Rightarrow-5=-25\text{x}$
$\Rightarrow\text{x}=\frac{5}{25}=\frac{1}{5}$

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MCQ 821 Mark

The value of the $\tan\Big(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\Big)=$

A
$\frac{6}{17}$
B
$\frac{7}{16}$
C
$\frac{16}{7}$
✓
None of these.

Answer

Correct option: D.

None of these.

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MCQ 831 Mark

The value of $\tan^{-1}\Big(\frac{1}{2}\Big)+\tan^{-1}\Big(\frac{1}{3}\Big)+\tan^{-1}\Big(\frac{7}{8}\Big)$ is:

A
$\tan^{-1}\Big(\frac{7}{8}\Big)$
B
$\cot^{-1}(15)$
✓
$\tan^{-1}(15)$
D
$\tan^{-1}\Big(\frac{25}{24}\Big)$

Answer

Correct option: C.

$\tan^{-1}(15)$

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MCQ 841 Mark

If $6\sin^{-1}(\text{x}^2-6\text{x}+8.5)=\pi,$ then $x$ is equal to:

A
$1$
✓
$2$
C
$3$
D
$8$

Answer

Correct option: B.

$2$

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MCQ 851 Mark

If $3\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)-4\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+2\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{\pi}{3}$ is equal to:

✓
$\frac{1}{\sqrt3}$
B
$-\frac{1}{\sqrt3}$
C
$\sqrt3$
D
$-\frac{\sqrt3}{4}$

Answer

Correct option: A.

$\frac{1}{\sqrt3}$

Let $\text{x}=\tan\text{y}$
Then,
$3\sin^{-1}\Big(\frac{\tan2\text{y}}{1+\tan^2\text{y}}\Big)-4\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)+2\tan^{-1}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)=\frac{\pi}{3}$
$\Rightarrow3\sin^{-1}(\sin2\text{y})-4\cos^{-1}(\cos2\text{y})+2\tan^{-1}(\tan2\text{y})=\frac{\pi}{3}$
$\Big[\because\ \sin2\text{y}=\Big(\frac{2\tan\text{y}}{1+\tan^2\text{y}}\Big),\cos2\text{y}=\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)$ and $\tan2\text{y}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)\Big]$
$\Rightarrow3\times2\text{y}-4\times2\text{y}+2\times2\text{y}=\frac{\pi}{3}$
$\Rightarrow6\text{y}-8\text{y}+4\text{y}=\frac{\pi}{3}$
$\Rightarrow2\text{y}=\frac{\pi}{3}$
$\Rightarrow\text{y}=\frac{\pi}{6}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$
$\big[\because\ \tan^{-1}\text{x}=\text{y}\big]$
$\Rightarrow\text{x}=\tan\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$

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MCQ 861 Mark

The domain of $\cos^{-1}\big(\text{x}^2-4\big)$ is:

A
$[3,5]$
B
$[-1,1]$
✓
$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
D
$\Big[-\sqrt5,-\sqrt3\Big]\cap\Big[\sqrt3,\sqrt5\Big]$

Answer

Correct option: C.

$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$

Let, $\cos^{-1}\big(\text{x}^2-4\big)=\text{y}$
$\Rightarrow\cos\text{y}=\text{x}^2-4$
$\Rightarrow-1\leq\text{x}^2-4\leq1$
$\Rightarrow3\leq\text{x}^2\leq5$
$\Rightarrow\pm\sqrt3\leq\text{x}\pm\sqrt5$
$\text{x}\in\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$

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MCQ 871 Mark

If $\text{x }\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),$ then the value of $\tan^{-1}\Big(\frac{\tan\text{x}}{4}\Big)+\tan^{-1}\Big(\frac{3\sin2\text{x}}{5+3\cos2\text{x}}\Big)$ is:

A
$\frac{\text{x}}{2}$
B
$2x$
C
$3x$
✓
$x$

Answer

Correct option: D.

$x$

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MCQ 881 Mark

$ \sin^{-1}\text{⁡x}+\cos^{1}\text{⁡x}= $

✓
$ \frac{\pi }{2}$
B
$\pi$
C
$\pi^3$
D
$2\pi $

Answer

Correct option: A.

$ \frac{\pi }{2}$

$ \sin-1\text{⁡x}+\cos-1\text{⁡x}=\pi ^2; \text{x} \in [-1,1] $

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MCQ 891 Mark

Solve: $\sin { \left( { \tan }^{ -1 }\text{x} \right) } ,\left| \text{x} \right| <1$ is equal to:

A
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
B
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
C
$\frac { \text{x} }{ \sqrt { 1-{ \text{x} }^{ 2 } } }$
✓
$\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$

Answer

Correct option: D.

$\frac { \text{x} }{ \sqrt { 1+{ \text{x} }^{ 2 } } }$

We need to find value of $ \sin (\tan^{-1}\text{x})$ Put $\text{y}=\tan^{-1}\text{x}$
$ \Rightarrow \displaystyle \tan {\text{ y} }$
$ \therefore \tan \text{y}=\frac {\sin \text{y}}{\cos \text{y}}$
$\Rightarrow \sin \text{y}=\frac{\tan \text{y}}{\sec \text{y}}$
$ \Rightarrow \sin { \text{y} } =\frac { \text{x} }{ \sqrt { 1+{\text{ x }}^{ 2 } } }$

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MCQ 901 Mark

The domain of the function defind by $\text{f(x)}=\sin^{-1}\sqrt{\text{x}-1}$ is:

✓
$[1, 2]$
B
$[-1, 1]$
C
$[0, 1]$
D
None of these.

Answer

Correct option: A.

$[1, 2]$

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MCQ 911 Mark

If $ \text{x}=\cos^{-1}(\cos 4): \text{y}=\sin^{-1}(\sin 3)$ then which of the following holds?

A
$x - y = 1$
B
$x + y + 1 = 0$
✓
$x + 2y = 2$
D
$y + x = 0$

Answer

Correct option: C.

$x + 2y = 2$

Given, $\text{x}=\cos^{-1}(\cos 4)$
$ = 2\pi - 4$
and $\text{y}=\sin^{-1}(\sin 3)\text{y}$
$ \pi - 3$
$ \tan(\text{x}+\text{y})$
$ =\tan(3\pi-4-3)$
$ =\tan(3\pi-7)$
$ =-\tan(7)$

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MCQ 921 Mark

If $ 3\cos ^{ -1 }{ \text{x} } +\sin ^{ -1 }{\text{ x} } =\pi $ then $x:$

A
$\frac { 4 }{ \sqrt { 2 } }$
B
$ -\frac { 1 }{ \sqrt { 2 } }$
✓
$\frac { 1 }{ \sqrt { 2 } }$
D
$\frac { 1 }{ \sqrt { 4 } }$

Answer

Correct option: C.

$\frac { 1 }{ \sqrt { 2 } }$

$ \sin^{-1}\text{x} +\cos^{-1}\text{x}=\frac{\pi}{2}$
$ =3\cos^{-1}\text{x}+\sin^{-1}\text{x}$
$ =2\cos^{1}\text{x}+\cos^{-1}\text{x}+\sin^{-1}\text{x}$
$=\sin^{−1}\text{x}=π$
$ = 2\cos^{-1}\text{x}+\frac{\pi}{2}$
$ =\pi=2\cos^{−1}\text{x}=\frac{\pi}{2}$
$= \cos^{-1}\text{x}=\frac{\pi}{4}\text{x}$
$=\cos (\frac{\pi}{4})=\frac{1}{\sqrt 2}$

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MCQ 931 Mark

If $\alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$ and $\beta=\tan^{-1}\Big(-\tan\frac{2\pi}{3}\Big),$ then:

✓
$4\alpha=3\beta$
B
$3\alpha=4\beta$
C
$\alpha-\beta=\frac{7\pi}{12}$
D
None of these

Answer

Correct option: A.

$4\alpha=3\beta$

We know that $\tan^{-1}(\tan\text{x})=\text{x}$
$\therefore\ \alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$
$=\tan^{-1}\Big\{\tan\Big(\pi+\frac{\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
and
$\beta=\tan^{-1}\Big\{-\tan\Big(\frac{2\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\pi-\frac{\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}$
$\therefore\ 4\alpha=\pi$
$3\beta=\pi$
$\therefore\ 4\alpha=3\beta$

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MCQ 941 Mark

Choose the correct answer from the given four options.If $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$ then $\cot^{-1}\text{x}+\cot^{-1}\text{y}$ equals to:

✓
$\frac{\pi}{5}$
B
$\frac{2\pi}{5}$
C
$\frac{3\pi}{5}$
D
$\pi$

Answer

Correct option: A.

$\frac{\pi}{5}$

We have, $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$
$\Rightarrow\ \frac{\pi}{2}-\cot^{-1}\text{x}+\frac{\pi}{2}-\cot^{-1}\text{y}=\frac{4\pi}{5}$
$\Rightarrow\ -(\cot^{-1}\text{x}+\cot^{-1}\text{y})=\frac{4\pi}{5}-\pi$
$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=-\Big(-\frac{\pi}{5}\Big)$
$\Rightarrow\ \cot^{-1}\text{x}+\cot^{-1}\text{y}=\frac{\pi}{5}$

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MCQ 951 Mark

The given graph is for which equation?

A
$\text{y}= \sin\text{x}$
✓
$\text{y} = \sin-1\text{x}$
C
$\text{y} = \text{cosec }\text{x}$
D
$\text{y} = \sec\text{x}$

Answer

Correct option: B.

$\text{y} = \sin-1\text{x}$

The following graph represents $2$ equations.

The pink curve is the graph of $\text{y} = \sin\text{x}$
The blue curve is the graph for $\text{y} = \sin^{-1}{\text{x}}$
This curve passes through the origin and approaches to infinity in both positive and negative axes.

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MCQ 961 Mark

What is the value of $ \cos (2 \cos^{-1} 0.8)\cos(2\cos−10.8)?$

A
$0.81$
✓
$0.56$
C
$0.48$
D
$0.28$

Answer

Correct option: B.

$0.56$

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MCQ 971 Mark

$\tan^{-1}1+\cos^{-1}\Big(\frac{-1}{2}\Big)+\sin^{-1}\Big(\frac{-1}{2}\Big)$

A
$\frac{2\pi}{3}$
✓
$\frac{3\pi}{4}$
C
$\frac{\pi}{2}$
D
$6\pi$

Answer

Correct option: B.

$\frac{3\pi}{4}$

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MCQ 981 Mark

The value of $\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$ is:

A
$\frac{1}{\sqrt2}$
B
$\frac{1}{\sqrt3}$
✓
$\frac{1}{2\sqrt2}$
D
$\frac{1}{3\sqrt3}$

Answer

Correct option: C.

$\frac{1}{2\sqrt2}$

$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
Let, $\sin^{-1}\frac{\sqrt{63}}{8}=\text{x}$
$\sin\text{x}=\frac{\sqrt{63}}{8}$
$\cos\text{x}\sqrt{1-\sin^2\text{x}}$
$\cos\text{x}=\sqrt{1-\frac{63}{64}}$
$\cos\text{x}=\frac{1}{8}$
Consider,
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
$=\sin\Big(\frac{1}{4}\text{x}\Big)$
$=\sqrt{\frac{1-\cos\frac{\text{x}}{2}}{2}}$ $\Big(\because \sin\text{x}=\frac{1-\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{\frac{1+\cos\text{x}}{2}}}{2}}$ $\Big(\because \cos\text{x}=\frac{1+\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{1-\frac{1}{8}}}{2}}$
$=\sqrt{\frac{1-\frac{3}{4}}{2}}$
$=\sqrt{\frac{1}{8}}$
$=\frac{1}{2\sqrt2}$

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MCQ 991 Mark

If the polynomial equation $\text{a}_0\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0\ n$ positive integer,has two different real roots $\alpha$ and $\beta,$ then between $\alpha$ and $\beta,$ the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ has:

A
Exactly one root.
B
Almost one root.
✓
At least one root.
D
No root.

Answer

Correct option: C.

At least one root.

We observe that, $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ is the derivative of the polynomial $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$
Polynomial function is continuous everywhere in $R$ and concequently derivative in $R.$
Therefore, $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0$ is continuous on $\alpha,\beta$ and derivative on $\alpha,\beta.$
Hence, it is satisfies the both the conditions of Rolle's theorem.
By algebric interpretation of Roll's theorem, we know that between any two roots of a function $f(x),$ there exists atleast one root of its derivative.
Hence, the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ will have atleast one root between $\alpha$ and $\beta.$

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MCQ 1001 Mark

If $x$ takes negative permissible value, then $\sin−1\text{x} $ is equal to:

A
$\cos^{-1}\sqrt{1-\text{x}^2}$
✓
$-\cos^{-1}\sqrt{1-\text{x}^2}$
C
$\cos^{-1}\sqrt{\text{x}^2-1}$
D
$\pi-\cos^{-1}\sqrt{1-\text{x}^2}$

Answer

Correct option: B.

$-\cos^{-1}\sqrt{1-\text{x}^2}$

$\sin^{-1}(\text{x})$
$-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big),$ for $x > 0$
Since $x$ takes negative permissible value.
$\sin^{-1}\text{x}=-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$

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MCQ 1011 Mark

The number of real values of $x$ satisfying the equation $ \tan^{-1}\left(\frac{\text{x}}{1-\text{x}^2}\right)+\tan^{-1}\left(\frac{1}{\text{x}^3}\right)=\frac{3\pi}{4}$, is?

A
$0$
✓
$1$
C
$2$
D
Infinitely many

Answer

Correct option: B.

$1$

$\tan ^{-1} \frac{\text{x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$
$= ^{-1} \frac{3\pi \text{ x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$
$ =−1\text{x}^4 + 1 -\text{x}^2 + \text{x}^3 - \text{x}^5-\text{x} = 0$
$= 0$ Real roots $= 11$

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MCQ 1021 Mark

If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta=$

A
$\pm\frac{\pi}{3}$
B
$\pm\frac{\pi}{4}$
✓
$\pm\frac{\pi}{6}$
D
None of these

Answer

Correct option: C.

$\pm\frac{\pi}{6}$

We have, $\tan^{-1}(\cot\theta)=2\theta$
$\Rightarrow\tan2\ \theta=\cot\theta$
$\Rightarrow\frac{2\tan\theta}{1-\tan^2\theta}=\frac{1}{\tan\theta}$
$\Rightarrow2\tan^2\theta=1-\tan^2\theta$
$\Rightarrow3\tan^2\theta=1$
$\Rightarrow\tan^2\theta=\frac{1}{3}$
$\Rightarrow\tan\theta=\pm\frac{1}{\sqrt3}$

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MCQ 1031 Mark

$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)$

✓
$\frac{\pi}{4}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{6}$
D
$\frac{\pi}{2}$

Answer

Correct option: A.

$\frac{\pi}{4}$

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MCQ 1041 Mark

The value of $\cos^{-1}\Big(\cos\Big(\frac{33\pi}{5}\Big)\Big)$ is:

✓
$\frac{3\pi}{5}$
B
$\frac{-3\pi}{5}$
C
$\frac{\pi}{10}$
D
$\frac{-\pi}{10}$

Answer

Correct option: A.

$\frac{3\pi}{5}$

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MCQ 1051 Mark

Find the value of $\sec^2(\tan^{-1}2)+\text{cosec}^2(\cot^{-1}3)$

A
$12$
B
$5$
✓
$15$
D
$9$

Answer

Correct option: C.

$15$

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MCQ 1061 Mark

If two angles of a triangle are $ \tan ^{ -1 }{ (2) }$ and $ \tan ^{ -1 }{ (3) }$, then the third angle is:

✓
$ \frac { \pi }{ 4 }$
B
$ \frac { \pi }{ 5 }$
C
$ \frac { \pi }{ 6 }$
D
$ \frac { \pi }{ 8 }$

Answer

Correct option: A.

$ \frac { \pi }{ 4 }$

Given two angles are $ \tan ^{ -1 }{ (2) }$ and $ \tan ^{ -1 }{ (3) }$.
Now$, (2) (3) > 1$
$= \tan ^{ -1 }{ (2) } +\tan ^{ -1 }{ (3) }$
$ =\pi +\tan ^{ -1 }{ \cfrac { 2+3 }{ 1-2\times 3 }}$
$ =\pi +\tan ^{ -1 }{ (-1) } $
$=\pi -\frac { \pi }{ 4 } $
$=\frac { 3\pi }{ 4 }$
Hence the third angle is $ \pi -\frac { 3\pi }{ 4 } =\frac { \pi }{ 4 }\pi $

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MCQ 1071 Mark

If $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8,$ then $x =$

A
$5$
✓
$\frac{1}{5}$
C
$\frac{5}{14}$
D
$\frac{14}{5}$

Answer

Correct option: B.

$\frac{1}{5}$

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MCQ 1081 Mark

What is the value of $ \cos^{-1}(-\text{x})$ for all $x$ belongs to $[-1, 1]\ ?$

A
$ \cos^{-1}(-\text{x})$
✓
$\pi- \cos^{-1}(-\text{x})$
C
$ π – \cos-1(-\text{x})$
D
$ π – \cos-1(+\text{x})$

Answer

Correct option: B.

$\pi- \cos^{-1}(-\text{x})$

Let, $ \theta = \cos-1(\text{-x})$
So, $ 0 \leq \theta \leq \pi $
$ \Rightarrow -\text{x} = \cos\theta $
$ \Rightarrow \text{x} = -\cos\theta $
$ \Rightarrow \text{x} = \cos-\theta $
Also, $ -\pi \leq -\theta \leq 0$
So, $ 0 \leq \pi -\theta \leq \pi $
$ \Rightarrow -\theta = \cos^{-1}(\text{x})$
$ \Rightarrow \theta = \cos^{-1}(\text{x})$
So, $\cos-1(\text{x}) = \pi – \theta $
$ \theta = \pi – \cos-1(\text{x})$
$ \Rightarrow \cos^{-1}(-\text{x}) = \pi – \cos^{-1}(\text{x})$

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MCQ 1091 Mark

Choose the correct answer from the given four options.If $|\text{x}|\leq1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:

✓
$4\tan^{-1}\text{x}$
B
$0$
C
$\frac{\pi}{2}$
D
$\pi$

Answer

Correct option: A.

$4\tan^{-1}\text{x}$

We have, $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Let $\text{x}=\tan\theta$
$\therefore\ 2\tan^{-1}\tan\theta+\sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta}$
$[\because\ \tan^{-1}(\tan\text{x})=\text{x}]$
$=2\theta+\sin^{-1}\sin2\theta$
$\Big[\because\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$=2\theta+2\theta$
$[\because\ \sin^{-1}=(\sin\text{x})=\text{x}]$
$=4\theta\ [\because\ \theta=\tan^{-1}\text{x}]$
$=4\tan^{-1}\text{x}$

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MCQ 1101 Mark

What is the value of $ {\sin}^{-1}(\sin 160^{\circ})?$

A
$160^{\circ}$
B
$70^{\circ}$
C
$-20^{\circ}$
✓
$20^{\circ}$

Answer

Correct option: D.

$20^{\circ}$

sinsin of an angle is positive in first and second quadrants.
$\Rightarrow \sin ^{ -1 }{ (\sin { { 160 }^{ \circ } } } )$
$\Rightarrow(\sin ^{ -1 }{ (\sin { { (180-20) }^{ \circ } } })=20^\circ$

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MCQ 1111 Mark

$\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}=$

A
$\pi$
B
$\frac{\pi}{2}$
✓
$\frac{\pi}{4}$
D
$\frac{3\pi}{4}$

Answer

Correct option: C.

$\frac{\pi}{4}$

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MCQ 1121 Mark

If $\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$ then $\text{f}\Big(\frac{8\pi}{9}\Big)=$

A
$\text{e}^{\frac{5\pi}{18}}$
✓
$\text{e}^{\frac{13\pi}{18}}$
C
$\text{e}^{\frac{-2\pi}{18}}$
D
None of these

Answer

Correct option: B.

$\text{e}^{\frac{13\pi}{18}}$

Given
$\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$
Then,
$\text{f}\Big(\frac{8\pi}{9}\Big)=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{8\pi}{9}+\frac{\pi}{3}\big)\big\}}$
$=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{11\pi}{9}\big)\big\}}=\text{e}^{\cos^{-1}\big\{\cos\frac{\pi}{2}+\frac{13\pi}{18}\big\}}$
$\Big[\because\ \cos\Big(\frac{\pi}{2}+\theta\Big)=\sin\theta\Big]$
$=\text{e}^{\cos^{-1}\big\{\cos\big(\frac{13\pi}{18}\big)\big\}}$
$=\text{e}^{\frac{13\pi}{18}}$

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MCQ 1131 Mark

If $\theta=\sin^{-1}\{\sin(-600^\circ)\},$ then one of the possible values of $\theta$ is:

✓
$\frac{\pi}{3}$
B
$\frac{\pi}{2}$
C
$\frac{2\pi}{3}$
D
$-\frac{2\pi}{3}$

Answer

Correct option: A.

$\frac{\pi}{3}$

$\theta=\sin^{-1}\{\sin(-600^\circ)\}$
$\theta=\sin^{-1}[\sin(-600^\circ)]$
$\theta=\sin^{-1}[-\sin(180^\circ\times3+60)]$
$\theta=\sin^{-1}[-\{-\sin(60^\circ)\}]$
$\theta=\sin^{-1}(\sin(60^\circ))$
$\theta=\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$
$\theta=\frac{\pi}{3}$

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MCQ 1141 Mark

$\tan\Big(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\Big)=$

A
$\text{x}$
B
$\frac{1}{\text{x}}$
C
$2\text{x}$
✓
$\frac{2}{\text{x}}$

Answer

Correct option: D.

$\frac{2}{\text{x}}$

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MCQ 1151 Mark

The value of $\tan^{-1}\Big(\frac{3}{4}\Big)+\tan^{-1}\Big(\frac{1}{7}\Big)$ is:

A
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{3\pi}{4}$
✓
$\frac{\pi}{4}$

Answer

Correct option: D.

$\frac{\pi}{4}$

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MCQ 1161 Mark

The value of $ \left( {{{\tan }^{ - 1}}\pi + {{\tan }^{ - 1}}\left( {\frac{1}{\pi }} \right)} \right) + {\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$ is equal to

A
$ \frac{{ - \pi }}{3}$
✓
$ \frac{{ \pi }}{6}$
C
$ \frac{{ 2 \pi }}{3}$
D
$ \pi$

Answer

Correct option: B.

$ \frac{{ \pi }}{6}$

$ \left( {{{\tan }^{ - 1}}\pi + {{\tan }^{ - 1}}\left( {\frac{1}{\pi }} \right)} \right) + {\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$
$=\tan^{-1}\pi+\cot^{-1}\pi+\tan^{-1}\sqrt 3-\sec^{-1}(-2)$
$ [\because \tan^{−1}\frac{1}{\text{y}}=\cot−1\text{y}]$
$ =\cfrac{\pi}{2}+\tan^{-1}\sqrt 3-\sec^{-1}(-2)\Big[\because \tan^{−1}x+\cot^{−1}\text{x}=\frac{\pi }{2}\Big]=\frac{\pi }{2}+\frac{\pi }{3}-\frac{2\pi }{3}$
$ =[\because \tan\frac{\pi }{3}=\sqrt{3};\sec\frac{2\pi }{3}=−2]$
$ =\frac{\pi }{2}−\frac{\pi }{3}$
$ =\frac{\pi}{6}$
None of the given options are correct.

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MCQ 1171 Mark

If $\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$ then, $\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)=$

✓
$\sqrt{\tan\theta}$
B
$\sqrt{\cot\theta}$
C
$\tan\theta$
D
$\cot\theta$

Answer

Correct option: A.

$\sqrt{\tan\theta}$

Let $\text{y}=\sqrt{\tan\theta}$
Then,
$\Rightarrow\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$
$\Rightarrow\text{u}=\cot^{-1}\text{y}-\tan^{-1}\text{y}$
$\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow2\tan^{-1}\text{y}=\frac{\pi}{2}-\text{u}$
$\Rightarrow\tan^{-1}\text{y}=\frac{\pi}{4}-\frac{\text{u}}{2}$
$\Rightarrow\text{y}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$
$\Rightarrow\sqrt{\tan\theta}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$
$\Big[\because\ \text{y}=\sqrt{\tan\theta}\Big]$

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MCQ 1181 Mark

If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then $x =$

A
$\frac{1}{2}$
✓
$\frac{\sqrt3}{2}$
C
$-\frac{1}{2}$
D
None of these

Answer

Correct option: B.

$\frac{\sqrt3}{2}$

We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\therefore \sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\frac{\pi}{2}-\cos^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow-2\cos^{-1}\text{x}=\frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow-2\cos^{-1}\text{x}=-\frac{\pi}{3}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{\sqrt3}{2}$

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MCQ 1191 Mark

The value of $\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]:$

✓
$ \frac { 1 }{ \sqrt { 5 } }$
B
$ \frac { 1 }{ \sqrt { 7 } }$
C
$ \frac { 1 }{ \sqrt { 8 } }$
D
$ \frac { 1 }{ \sqrt { 9 } }$

Answer

Correct option: A.

$ \frac { 1 }{ \sqrt { 5 } }$

We have, $\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$
Put $ \cos ^{ -1 }{ \frac { 2 }{ 3 } } =θ$
$ \Rightarrow \cos { \theta } =\frac { 2 }{ 3 }$
$\therefore\tan\bigg[\frac{1}{2}\cos^{-1}\Big(\frac{2}{3}\Big)\bigg]$
$ \therefore \ \tan\frac{\theta}{2}=\sqrt {\frac {1-\cos {\theta}}{1+\cos{\theta }}}$
$=\sqrt {\frac {1-\frac{2}{3} }{1+{\frac{ 2}{3}}}}$
$=\sqrt { \frac { \frac{ 1 }{ 3 } }{ \frac{5}{3} } } $
$=\frac { 1 }{ \sqrt { 5 } }$

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MCQ 1201 Mark

Choose the correct answer from the given four options. If $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$ then $x$ is equal to:

A
$\frac{1}{5}$
✓
$\frac{2}{5}$
C
$0$
D
$1$

Answer

Correct option: B.

$\frac{2}{5}$

We have, $\cos\Big(\sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}\Big)=0$
$\Rightarrow \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\cos^{-1}0$
$\Rightarrow \sin^{-1}\frac{2}{5}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\frac{2}{5}$
$\Rightarrow \cos^{-1}\text{x}=\cos^{-1}\frac{2}{5}$
$\Big(\because \cos^{-1}\text{x}+\sin^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\therefore \text{x}=\frac{2}{5}$

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MCQ 1211 Mark

The value of $ \cos { \left( \tan ^{ -1 }{ \tan { 4 } } \right) }$ is$-$

A
$ \frac { 1 }{ \sqrt { 17 } }$
B
$ \frac { 1 }{ \sqrt {- 17 } }$
C
$ \frac { 1 }{ \sqrt {- 14 } }$
✓
$ -\cos 4$

Answer

Correct option: D.

$ -\cos 4$

As for $ \displaystyle \tan ^{ -1 }{ \text{x} }; \text{x}\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$
$\cos(\tan^{−1}(\tan4))$
$=\cos(\tan^{−1}(\tan(π−4))$
$ =\cos(π−4)$
$=−\cos4$

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MCQ 1221 Mark

If $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{A},$ then $A$ is equal to:

A
$\text{x}-\text{y}$
B
$\text{x}+\text{y}$
✓
$\frac{\text{x}-\text{y}}{1+\text{xy}}$
D
$\frac{\text{x}+\text{y}}{1-\text{xy}}$

Answer

Correct option: C.

$\frac{\text{x}-\text{y}}{1+\text{xy}}$

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MCQ 1231 Mark

Find the value of $ {\sin ^{ - 1}}\left( 1 \right):$

A
$ \dfrac{\pi}{7}$
B
$ \dfrac{\pi}{6}$
C
$ \dfrac{\pi}{4}$
✓
$ \dfrac{\pi}{2}$

Answer

Correct option: D.

$ \dfrac{\pi}{2}$

Value of $\sin^{-1}(1)\sin\text{x}$ is in vertible form $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ in this range only $\sin\frac{\pi}{2}=1\sin^{-1}(1)\frac{\pi}{2}.$

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MCQ 1241 Mark

Choose the correct answer from the given four options. The value of the expression $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$ is:

A
$\frac{\pi}{6}$
✓
$\frac{5\pi}{6}$
C
$\frac{7\pi}{6}$
D
$1$

Answer

Correct option: B.

$\frac{5\pi}{6}$

We have, $2\sec^{-1}2+\sin^{-1}\Big(\frac{1}{2}\Big)$
$=2\sec^{-1}\sec\frac{\pi}{3}+\sin^{-1}\sin\frac{\pi}{6}$
$=2\frac{\pi}{3}+\frac{\pi}{6}$
$[\because \sec^{-1}(\sec\text{x})=\text{x} $ and $ \sin^{-1}(\sin\text{x})=\text{x}]$
$=\frac{4\pi+\pi}{6}$
$=\frac{5\pi}{6}$

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MCQ 1251 Mark

The equation $2\cos^{-1}\text{x}+\sin^{-1}\text{x}=\frac{11\pi}{6}$ has:

✓
No solution.
B
Only one solution.
C
Two solutions.
D
Three solutions.

Answer

Correct option: A.

No solution.

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MCQ 1261 Mark

If $ \cos^{-1}\left (\frac {1 - \text{x}^{2}}{1 + \text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right )=\frac{\pi}{2}$, where $xy < 1,$ then:

A
$x - y - xy = 1$
B
$x - y + xy = 1$
C
$x + y - xy = 1$
✓
$x + y + xy = 1$

Answer

Correct option: D.

$x + y + xy = 1$

Given, $ \cos^{-1} \left (\frac {1 - \text{x}^{2}}{1 +\text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right ) = \frac {\pi}{2}$
$ \Rightarrow \tan^{-1} \left (\frac {\text{x} + \text{y}}{1 - \text{xy}}\right ) = \frac {\pi}{4}$
$ \Rightarrow \frac {\text{x} + \text{y}}{1 -\text{ xy}} = \tan \frac {\pi}{4}$
$\Rightarrow \text{x} + \text{y} = 1 - \text{xy} = \text{x} + \text{y} + \text{xy}$

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MCQ 1271 Mark

The value of $\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$ is:

A
$\frac{\pi}{2}$
B
$\frac{5\pi}{3}$
C
$\frac{10\pi}{3}$
✓
$0$

Answer

Correct option: D.

$0$

We have
$\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$
$=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{\sin\Big(2\pi-\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{-\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}-\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}-\frac{\pi}{3}$
$=0$

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MCQ 1281 Mark

If $\tan^{-1}\Big(\frac{\text{a}}{\text{x}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{x}}\Big)=\frac{\pi}{2},$ then $x$ is equal to:

✓
$\sqrt{\text{ab}}$
B
$\sqrt{2\ \text{ab}}$
C
$2\ \text{ab}$
D
$\text{ab}$

Answer

Correct option: A.

$\sqrt{\text{ab}}$

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MCQ 1291 Mark

The value of $\sin\big(2\big(\tan^{-1}0.75\big)\big)$ is equal to:

A
$0.75$
B
$1.5$
✓
$0.96$
D
$\sin ^{-1} 1.5$

Answer

Correct option: C.

$0.96$

$\sin\big(2\big(\tan^{-1}0.75\big)\big)$
$=\sin\big(2\tan^{-1}0.75\big)$
$=\sin\Big(\sin^{-1}\frac{2\times0.75}{1+(0.75)^2}\Big)$
$=\sin\big(\sin^{-1}0.96\big)$
$=0 .96$
Hence, the correct answer is option $(c).$

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MCQ 1301 Mark

Choose the correct answer from the given four options. If $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi,$ then $x$ equals:

A
$0$
✓
$1$
C
$-1$
D
$\frac{1}{2}$

Answer

Correct option: B.

$1$

We have, $3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+(\tan^{-1}\text{x}+\cot^{-1}\text{x})=\pi$
$\Rightarrow 2\tan^{-1}\text{x}+\frac{\pi}{2}=\pi$
$\Big(\because \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\Rightarrow 2\tan^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow \tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow \text{x}=1$

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MCQ 1311 Mark

Choose the correct answer from the given four options.The domain of the function defined by $\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$ is:

✓
$[1, 2]$
B
$[-1, 1]$
C
$[0, 1]$
D
none of these.

Answer

Correct option: A.

$[1, 2]$

$\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$
$\Rightarrow\ 0\leq\text{x}-1\leq1\ [\because\ \sqrt{\text{x}-1}\geq0$ and $-1\leq\sqrt{\text{x}-1}\leq1]$
$\Rightarrow\ 1\leq\text{x}\leq2$
$\therefore\ \text{x}\in[1,2]$

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MCQ 1321 Mark

The number of solution of the equation $ 1+\text{x}^{2}+2\text{x}\:\sin \left ( \cos^{-1}\text{y} \right )= 0$ is:

✓
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct option: A.

$1$

Given, $ 1+\text{x}^2+2\text{x}(\sin(\cos^{-1}(\text{y})))=0$
$ 1+\text{x}^2+2\text{x}(\sin(\sin^{-1}(\sqrt{1-\text{y}^2})))=0$
$ 1+\text{x}^2+2\text{x}(\sqrt{1-\text{y}^2})=0$
$ 2\text{x}(\sqrt{1-\text{y}^2})=-(1+\text{x}^2)$
$ 4\text{x}^2(1-\text{y}^2)=1+\text{x}^4+2\text{x}^2$
$ 4\text{x}^2-4\text{x}^2\text{y}^2=1+\text{x}^4+2\text{x}^2$
$ -4\text{x}^{2}(\text{y}^2)=(1-\text{x}^2)^{2}$
Hence solution will be $x = 1$ and $\text{y}=\frac{1}{2}$

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MCQ 1331 Mark

$\sin^{-1}(1-\text{x})-2\sin^{-1}\text{x}=\frac{\pi}{2}$

✓
$0$
B
$\frac{1}{2}$
C
$0,\frac{1}{2}$
D
$-\frac{1}{2}$

Answer

Correct option: A.

$0$

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MCQ 1341 Mark

$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=$

✓
$7$
B
$6$
C
$5$
D
None of these

Answer

Correct option: A.

$7$

Let $2\cot^{-1}3=\text{y}$
Then, $\cot\frac{\text{y}}{2}=3$
$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)$
$=\cot\Big(\frac{\pi}{4}-\text{y}\Big)$
$=\frac{\cot\frac{\pi}{4}\cot\text{y}+1}{\cot\text{y}-\cot\frac{\pi}{4}}$
$=\frac{\cot\text{y}+1}{\cot\text{y}-1}$
$=\frac{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}+1}{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}-1}$
$=\frac{\cot^2\frac{\text{y}}{2}+2\cot\frac{\text{y}}{2}-1}{\cot^2\frac{\text{y}}{2}-2\cot\frac{\text{y}}{2}-1}$
$=\frac{9+6-1}{9-6-1}$
$=7$

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MCQ 1351 Mark

If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta$ is equal to:

A
$\frac{\pi}{3 }$
B
$\frac{\pi}{4}$
✓
$\frac{\pi}{6}$
D
None of these.

Answer

Correct option: C.

$\frac{\pi}{6}$

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MCQ 1361 Mark

The positive integral solution of the equation $\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$ is:

✓
$x = 1, y = 2$
B
$x = 2, y = 1$
C
$x = 3, y = 2$
D
$x = -2, y = -1$

Answer

Correct option: A.

$x = 1, y = 2$

We have,
$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\begin{bmatrix}\frac{\sqrt{1-\bigg(\frac{\text{y}}{\sqrt{1+\text{y}^2}}\bigg)^2}}{\frac{\text{y}}{\sqrt{1+\text{y}^2}}}\end{bmatrix}=\tan^{-1}\begin{bmatrix}\frac{\frac{3}{\sqrt{10}}}{\sqrt{1-\Big(\frac{3}{\sqrt{10}}\Big)^2}}\end{bmatrix}$

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MCQ 1371 Mark

$\sin^{−1}\text{x}+\sin^{−1}\frac{1}{\text{x}}+\cos^{−1}\text{x}+\cos^{−1}\frac{1}{\text{x}}=$

✓
$\pi$
B
$2\pi$
C
$ \cfrac{3\pi}{2}$
D
None of these

Answer

Correct option: A.

$\pi$

We know, $ \displaystyle \sin ^{ -1 }{ \theta } +\cos ^{ -1 }{ \theta } =\frac { \pi }{ 2 }$
$\therefore \sin ^{ -1 }{ \text{x} } +\sin ^{ -1 }{ \frac { 1 }{ \text{x} } } +\cos ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \frac { 1 }{ \text{x} } }$
$ \displaystyle =\frac { \pi }{ 2 } +\frac { \pi }{ 2 } $
$=\pi$

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MCQ 1381 Mark

Simplify $ {\cot ^{ - 1}}\frac{1}{{\sqrt {{\text{x}^2} - 1} }} $ for $\text{ x} < - 1$:

A
$ \cos ^{-1}\text{x}$
✓
$ \sec ^{-1}\text{x}$
C
$ \text{cosec} ^{-1}\text{x}$
D
$ \tan ^{-1}\text{x}$

Answer

Correct option: B.

$ \sec ^{-1}\text{x}$

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MCQ 1391 Mark

$\sin[\cot^{-1}\{\cos(\tan^{-1}\text{x})\}]=$

✓
$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$
B
$\sqrt{\frac{\text{x}^2-1}{\text{x}^2-2}}$
C
$\sqrt{\frac{\text{x}-1}{\text{x}-2}}$
D
$\sqrt{\frac{\text{x}+1}{\text{x}+2}}$

Answer

Correct option: A.

$\sqrt{\frac{\text{x}^2+1}{\text{x}^2+2}}$

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MCQ 1401 Mark

The number of solutions for the equation $ 2\sin ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x}+1 } } +\cos ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x} } } =\frac { 3\pi }{ 2 }$ is:

A
$1$
✓
$2$
C
$3$
D
Infinite

Answer

Correct option: B.

$2$

$ 2\sin ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x}+1 } } +\cos ^{ -1 }{ \sqrt { { \text{x} }^{ 2 }- \text{x} } } =\frac { 3\pi }{ 2 }$
For existence of domain of
$\sin^{-1}\sqrt{\text{x} ^2-\text{x}+1}-1\leq \sqrt{\text{x}^2-\text{x}}+1\leq 1$
$ 0 \leq \text{x}^2-\text{x}+1\leq 1$
$\text{x}\in [0,1]$
For $\cos^{-1}\sqrt{\text{x}^2-\text{x}0}\leq \text{x}^2-\text{x}\leq 1$
$ \Rightarrow \text{x}^2−\text{x}\geq 0$
$ \Rightarrow \text{x}−\text{x}\geq 0$
$\text{x}\in [−\infty ,0] \cup [1,\infty ]$ Only two points are common in their domains
i.e.$0$ and $1$ which also satisfies the given equation.
So option $B$ is correct.

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MCQ 1411 Mark

$\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha,$ then $\frac{\text{x}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha+\frac{\text{y}^2}{\text{b}^2}=$

✓
$\sin^2\alpha$
B
$\cos^2\alpha$
C
$\tan^2\alpha$
D
$\cot^2\alpha$

Answer

Correct option: A.

$\sin^2\alpha$

$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
Consider, $\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha$
$\Rightarrow\cos^{-1}\bigg(\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}=\cos\alpha$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\cos\alpha=\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}$
Squaring on both sides,
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\Big(1-\frac{\text{x}^2}{\text{a}^2}\Big)\Big(1-\frac{\text{y}^2}{\text{a}^2}\Big)$
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{a}^2}+\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\cos^2\alpha$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\sin^2\alpha$

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MCQ 1421 Mark

$\sin−10 $ is equal to:

✓
$0$
B
$ \dfrac{\pi }{6}$
C
$ \dfrac{\pi}{2}$
D
$ \dfrac{\pi}{3}$

Answer

Correct option: A.

$0$

As we know that $\sin{0} = 0\sin0=0$
$\Rightarrow 0 = \sin^{-1}{\left( 0 \right)}$
Hence the value of $ \sin^{-1}{\left( 0 \right)}$ is $0.$

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MCQ 1431 Mark

If $ \alpha \leq 2\sin^{-1} \text{x} + \cos^{-1} \text{x}\leq \betaα$ then:

A
$ \alpha = 0, \beta = \pi(2)$
B
$ \alpha = 0, \beta = 2\pi$
✓
$ \alpha = 0, \beta = \pi$
D
$ \alpha = 0, \beta = 5\pi$

Answer

Correct option: C.

$ \alpha = 0, \beta = \pi$

We know that, $ -\frac{\pi}{2}\le \sin^{-1}\text{x}\le \frac{\pi}{2}$
Now add $ \frac{\pi}{2}$
each sides $ \frac{\pi}{2}-\frac{\pi}{2}\le \sin^{-1}\text{x}+\frac{\pi}{2}\le \frac{\pi}{2}+\frac{\pi}{2}$
$ \Rightarrow 0\le \sin^{-1}+\dfrac{\pi}{2}\le \pi$
$ \Rightarrow 0\le \sin^{-1}+(\sin^{-1}\text{x}+\cos^{-1}\text{x})\le \pi$ use the identity
$ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$ \therefore 0\le 2\sin^{-1}\text{x}+\cos^{-1}\text{x}\le \pi$
Hence $ \alpha=0, \beta=\pi$

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MCQ 1441 Mark

If $x > 1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:

A
$4\tan^{-1}\text{x}$
B
$0$
✓
$\frac{\pi}{2}$
D
$\pi$

Answer

Correct option: C.

$\frac{\pi}{2}$

$2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=2\tan^{-1}\text{x}+2\tan^{-1}\text{x}$
$\Big[\because \sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$=4\tan^{-1}\text{x}$

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MCQ 1451 Mark

Choose the correct answer from the given four options.The number of real solutions of the equation $\sqrt{1+\cos2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x})$ in $\Big[\frac{\pi}{2},\pi\Big]$ is:

✓
$0$
B
$1$
C
$2$
D
$\infty$

Answer

Correct option: A.

$0$

We have $\sqrt{1+\cos2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x}),$
$\text{x}\in\Big[\frac{\pi}{2},\pi\Big]$
$\Rightarrow\ \sqrt{2\cos^2\text{x}}=\sqrt{2}\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \sqrt{2}\cos\text{x}=\sqrt{2}\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \cos\text{x}=\cos^{-1}(\cos\text{x})$
$\Rightarrow\ \cos\text{x}=\text{x}$
$[\because\ \cos^{-1}(\cos\text{x})=\text{x}]$
For $\text{x}\in\Big[\frac{\pi}{2},\pi\Big],\ \cos\text{x}\leq0$
$\therefore \cos x = x$ is not possible for any value of $x.$

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MCQ 1461 Mark

$\cos^{-1}\Big(\frac{1}{2}\Big)$

A
$-\frac{\pi}{3}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$\frac{2\pi}{3}$

Answer

Correct option: B.

$\frac{\pi}{3}$

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MCQ 1471 Mark

If $4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of $x$ is:

A
$\frac{3}{2}$
B
$\frac{1}{\sqrt2}$
✓
$\frac{\sqrt3}{2}$
D
$\frac{2}{\sqrt3}$

Answer

Correct option: C.

$\frac{\sqrt3}{2}$

We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi$
$\Rightarrow4\cos^{-1}\text{x}+\frac{\pi}{2}-\cos^{-1}\text{x}=\pi $
$\Rightarrow3\cos^{-1}\text{x}=\pi-\frac{\pi}{2}$
$\Rightarrow3\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$

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MCQ 1481 Mark

If $\sin\Big(\sin^{-1}\frac{1}{5}+\cos^{-1}\text{x}\Big)=1,$ then the value of $x$ is:

A
$-1$
B
$\frac{2}{5}$
C
$\frac{1}{3}$
✓
$\frac{1}{5}$

Answer

Correct option: D.

$\frac{1}{5}$

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MCQ 1491 Mark

Find the value of $x$ if $ \sin (\text{arc} \sin \text{x}) = \frac {\sqrt {2}}{4}:$

✓
$ \frac {\sqrt {2}}{4}$
B
$ \frac {\sqrt {2}}{6}$
C
$ \frac {\sqrt {6}}{4}$
D
$ \frac {\sqrt {2}}{3}$

Answer

Correct option: A.

$ \frac {\sqrt {2}}{4}$

Given, $ \sin \text{arc} \sin { \text{x} } =\frac { \sqrt { 2 } }{ 4 } =\frac { 1 }{ 2\sqrt { 2 } }$
$ \therefore \text{arc}\sin { \text{x} } =\text{arc}\sin \left (\frac {1}{2\sqrt2}\right)=0.36136$
$ \therefore \text{x}=\sin(0.36136)=\frac { 1 }{ 2\sqrt { 2 } }$

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MCQ 1501 Mark

ind the value of $\text{cot}\text{(tan}^1\text{a}+\text{cot}^1\text{a}).$

✓
$0$
B
$−1$
C
$2$
D
$1$

Answer

Correct option: A.

$0$

We know,
$\tan^1\text{a}+\cot^{-1}\text{a}=\frac{\pi}{2}$
Therefore,
$\cot(\tan^{−1}\text{a}+\cot^{−1}a)=\cot\frac{\pi}{2}=0$

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MCQ 1511 Mark

Choose the correct answer from the given four options.The value of $\cot\Big[\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$ is:

A
$\frac{25}{24}$
B
$\frac{25}{7}$
C
$\frac{24}{25}$
✓
$\frac{7}{24}$

Answer

Correct option: D.

$\frac{7}{24}$

$\cot\Big[\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$
$=\cot\Big(\cot^{-1}\frac{7}{24}\Big)$
$=\frac{7}{24}$

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MCQ 1521 Mark

The value of $\sin ^{ -1 }{ \left( \cos { \frac { 53\pi }{ 5 } } \right) }=\sin ^{ -1 }{ \left( \cos { \frac { 50\pi+3\pi }{ 5 } } \right) }:$

A
$ \frac {- \pi }{ 1 }$
B
$ \frac {- \pi }{ 7 }$
C
$ \frac { \pi }{ 10 }$
✓
$ \frac {- \pi }{ 10 }$

Answer

Correct option: D.

$ \frac {- \pi }{ 10 }$

We have: $\sin ^{ -1 }{ \left( \cos { \frac { 53\pi }{ 5 } } \right) }=\sin ^{ -1 }{ \left( \cos { \frac { 50\pi+3\pi }{ 5 } } \right) }$
$ =\sin ^{ -1 }{ \left( \cos \left({ \frac { 50\pi}{5}+\frac{3\pi }{ 5 } }\right) \right) }$
$ =\sin ^{ -1 }{ \left( \cos \left({ 10\pi+\frac{3\pi }{ 5 } }\right) \right) }$
$ =\sin ^{ -1 }{ \left( \cos \left({ \frac{3\pi }{ 5 } }\right) \right) }, [\because \cos(2\text{n}\pi+\theta)=\cos\theta, n\in \text{Z}]$
$ =\sin ^{ -1 }{ \left( \sin \left({ \frac{\pi}{2}-\frac{3\pi }{ 5 } }\right) \right) },[\because \sin(2\pi −\theta )=\cos\theta ]$
$ =\sin ^{ -1 }{ \left( \sin \left(-\frac{\pi}{10 }\right) \right) } =−\frac{\pi}{10}$
Note $ \sin^{-1}(\sin \theta)=\theta \text{ if} -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$

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MCQ 1531 Mark

If $\tan ^{-1} x=y$, then

A
-1 < y<1
B
$\frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$
C
$\frac{-\pi}{2}$ < y < $\frac{\pi}{2}$
D
$y \in\left\{\frac{-\pi}{2}, \frac{\pi}{2}\right\}$

Answer

Range of $\tan ^{-1} x=\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

$\therefore \frac{-\pi}{2}$< y < $\frac{\pi}{2}$

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MCQ 1541 Mark

Simplest form of
$\tan ^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)$,$\pi$< x<$\frac{3 \pi}{2}$ is

A
$\frac{\pi}{4}-\frac{x}{2}$
B
$\frac{3 \pi}{2}-\frac{x}{2}$
C
$-\frac{x}{2}$
D
$\pi-\frac{x}{2}$

Answer

We have,
$\tan ^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)$,$\pi$<x<$\frac{3 \pi}{2}$
$=\tan ^{-1}\left(\frac{\left|\sqrt{2} \cos \frac{x}{2}\right|+\left|\sqrt{2} \sin \frac{x}{2}\right|}{\left|\sqrt{2} \cos \frac{x}{2}\right|-\left|\sqrt{2} \sin \frac{x}{2}\right|}\right)$
$=\tan ^{-1}\left(\frac{-\sqrt{2} \cos \frac{x}{2}+\sqrt{2} \sin \frac{x}{2}}{-\sqrt{2} \cos \frac{x}{2}-\sqrt{2} \sin \frac{x}{2}}\right)$ $\left(\because \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}\right)$
$=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)=\frac{\pi}{4}-\frac{x}{2}$

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MCQ 1551 Mark

$\sin \left(\tan ^{-1} x\right)$, where $|x|<1$, is equal to

A
$\frac{x}{\sqrt{1-x^2}}$
B
$\frac{1}{\sqrt{1-x^2}}$
C
$\frac{1}{\sqrt{1+x^2}}$
D
$\frac{x}{\sqrt{1+x^2}}$

Answer

We have, $\sin \left(\tan ^{-1} x\right)$
Let $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta \Rightarrow \sin \theta=\frac{x}{\sqrt{x^2+1}}$
$\therefore \quad \sin \left(\tan ^{-1} x\right)=\sin \theta=\frac{x}{\sqrt{x^2+1}}$

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MCQ 1561 Mark

$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ is equal to

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$-1$
✓
$1$

Answer

Correct option: D.

$1$

We have,
$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right]$
$=\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]$
$=\sin \left[\frac{\pi}{3}+\frac{\pi}{6}\right]$
$=\sin \left(\frac{\pi}{2}\right)$
$=1$

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MCQ 1571 Mark

The value of $\sin ^{-1}\left(\cos \frac{13 \pi}{5}\right)$ is

A
$-\frac{3 \pi}{5}$
✓
$-\frac{\pi}{10}$
C
$\frac{3 \pi}{5}$
D
$\frac{\pi}{10}$

Answer

Correct option: B.

$-\frac{\pi}{10}$

$\text {We have, } \sin ^{-1}\left(\cos \frac{13 \pi}{5}\right)$
$=\sin ^{-1}\left[\cos \left(2 \pi+\frac{3 \pi}{5}\right)\right]$
$=\sin ^{-1}\left[\cos \frac{3 \pi}{5}\right]$
$=\sin ^{-1}\left[\cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right]$
$=\sin ^{-1}\left(-\sin \frac{\pi}{10}\right)$
$=-\sin ^{-1}\left(\sin \frac{\pi}{10}\right)$
$=-\frac{\pi}{10}$

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MCQ 1581 Mark

If $f(x)=|\cos x|$, then $f\left(\frac{3 \pi}{4}\right)$ is

A
$1$
B
$-1$
C
$\frac{-1}{\sqrt{2}}$
✓
$\frac{1}{\sqrt{2}}$

Answer

Correct option: D.

$\frac{1}{\sqrt{2}}$

$f(x)=|\cos x|$
At$\frac{\pi}{2} < x < \pi, \cos x < 0$
$\therefore|\cos x|=-\cos x$
$\Rightarrow f(x)=-\cos x$
$\therefore f\left(\frac{3 \pi}{4}\right)$
$=-\cos \left(\frac{3 \pi}{4}\right)$
$=-\cos \left(\pi-\frac{\pi}{4}\right)$
$=\cos \frac{\pi}{4}$
$=\frac{1}{\sqrt{2}} [\because \cos (\pi-\theta)=-\cos \theta]$

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MCQ 1591 Mark

$\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]$ is equal to

A
1
B
$\frac{1}{2}$
C
$\frac{1}{3}$
D
$\frac{1}{4}$

Answer

We have,
$\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]=\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right]=\sin \left[\frac{\pi}{3}+\frac{\pi}{6}\right]$
$=\sin \left(\frac{\pi}{2}\right)=1$

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MCQ 1601 Mark

The principal value of $\tan ^{-1}\left(\tan \frac{3 \pi}{5}\right)$ is

A
$\frac{2 \pi}{5}$
B
$\frac{-2 \pi}{5}$
✓
$\frac{3 \pi}{5}$
D
$\frac{-3 \pi}{5}$

Answer

Correct option: C.

$\frac{3 \pi}{5}$

We have, $\tan ^{-1}\left(\tan \frac{3 \pi}{5}\right)$
We know that the range of $\tan ^{-1} x$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
$\therefore \tan ^{-1}\left(\tan \frac{3 \pi}{5}\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{2 \pi}{5}\right)\right)$
$=\tan ^{-1}\left[-\tan \left(\frac{2 \pi}{5}\right)\right] \quad[\because \tan (\pi-\theta)=\tan \theta]$
$=-\tan ^{-1}\left[\tan \left(\frac{2 \pi}{5}\right)\right]=-\frac{2 \pi}{5} \quad\left[\because \tan ^{-1}(\tan \theta)=\theta\right]$

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MCQ 1611 Mark

$\tan ^{-1} 3+\tan ^{-1} \lambda=\tan ^{-1}\left(\frac{3+\lambda}{1-3 \lambda}\right)$ is valid for what values of $\lambda ?$

A
$\lambda \in\left(-\frac{1}{3}, \frac{1}{3}\right)$
B
$\lambda > \frac{1}{3}$
✓
$\lambda < \frac{1}{3}$
D
All real values of $\lambda$

Answer

Correct option: C.

$\lambda < \frac{1}{3}$

$\text {Given, } \tan ^{-1} 3+\tan ^{-1} \lambda$
$=\tan ^{-1}\left(\frac{3+\lambda}{1-3 \lambda}\right)$
$\tan ^{-1} 3+\tan ^{-1} \lambda$
$=\tan ^{-1}\left(\frac{3+\lambda}{1-3 \lambda}\right) \text { for } 3 \lambda < 1$
$\therefore 3 \lambda < 1$
$\Rightarrow \lambda < \frac{1}{3}$

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MCQ 1621 Mark

The principal value of $\cot ^{-1}(-\sqrt{3})$ is

A
$-\frac{\pi}{6}$
B
$\frac{\pi}{6}$
C
$\frac{2 \pi}{3}$
D
$\frac{5 \pi}{6}$

Answer

We know that $\cot ^{-1}(x) \in(0, \pi)$
$\begin{array}{ll}
\cot ^{-1}(-\sqrt{3})=\cot ^{-1}\left(-\cot \frac{\pi}{6}\right) & \\
=\cot ^{-1}\left[\cot \left(\pi-\frac{\pi}{6}\right)\right] & {[\because \cot (\pi-\theta)=-\cot \theta]} \\
=\cot ^{-1}\left[\cot \left(\frac{5 \pi}{6}\right)\right]=\frac{5 \pi}{6} & {\left[\because \cot ^{-1}[\cot \theta]=\theta\right]}
\end{array}
$Thus, the principal value of $\cot ^{-1}(-\sqrt{3})$ is $\frac{5 \pi}{6}$.

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MCQ 1631 Mark

The principal solution of $\sin ^{-1}\left(\sin \left(\frac{5 \pi}{3}\right)\right)$ is

A
$\frac{4 \pi}{3}$
B
$\frac{5 \pi}{3}$
C
$\frac{-5 \pi}{3}$
✓
$\frac{-\pi}{3}$

Answer

Correct option: D.

$\frac{-\pi}{3}$

$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\sin ^{-1}\left(\sin \left(\frac{5 \pi}{3}\right)\right)$
$=\sin ^{-1}\left(\sin \left(2 \pi-\frac{\pi}{3}\right)\right)$
$=\sin ^{-1}\left(\sin \left(\frac{-\pi}{3}\right)\right)$
$=\frac{-\pi}{3}$

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MCQ 1641 Mark

The principal solution of $\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ is

A
$\frac{7 \pi}{6}$
✓
$\frac{\pi}{6}$
C
$\frac{-\pi}{6}$
D
$\frac{-5 \pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{6}$

$\text { (b) : In }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right),$
$\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{6}\right)\right)$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$

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MCQ 1651 Mark

The principal solution of $\cos ^{-1}\left(\cos \left(\frac{9 \pi}{4}\right)\right)$ is

A
$\frac{7 \pi}{4}$
B
$\frac{-\pi}{4}$
C
$\frac{9 \pi}{4}$
✓
$\frac{\pi}{4}$

Answer

Correct option: D.

$\frac{\pi}{4}$

$\text { (d): In }[0, \pi]$
$\cos ^{-1}\left(\cos \left(\frac{9 \pi}{4}\right)\right)=\cos ^{-1}\left(\cos \left(2 \pi+\frac{\pi}{4}\right)\right)$
$=\left(\cos ^{-1}\left(\cos \frac{\pi}{4}\right)\right)=\frac{\pi}{4}$

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MCQ 1661 Mark

If $\tan ^{-1}(\cot \theta)=2 \theta$, then $\theta$ is equal to

A
$\frac{\pi}{3}$
B
$\frac{\pi}{4}$
✓
$\frac{\pi}{6}$
D
None of these

Answer

Correct option: C.

$\frac{\pi}{6}$

$\text { (c) }: \tan ^{-1}(\cot \theta)=2 \theta$
$\Rightarrow \cot \theta=\tan 2 \theta$
$\Rightarrow \cot \theta=\cot \left(\frac{\pi}{2}-2 \theta\right)$
$\Rightarrow \theta=\frac{\pi}{2}-2 \theta$
$\Rightarrow 3 \theta=\frac{\pi}{2}$
$\Rightarrow \theta=\frac{\pi}{6}$

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MCQ 1671 Mark

$\cos ^{-1}\left[\cos \left(2 \cot ^{-1}(\sqrt{3})\right)\right]=$

A
$\frac{2 \pi}{3}$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{4}$
✓
$\frac{\pi}{3}$

Answer

Correct option: D.

$\frac{\pi}{3}$

We have, $\cos ^{-1}\left[\cos \left(2 \cot ^{-1}(\sqrt{3})\right)\right]$
$=\cos ^{-1}\left[\cos 2\left(\frac{\pi}{6}\right)\right]$
$=\cos ^{-1}\left(\cos \left(\frac{\pi}{3}\right)\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

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MCQ 1681 Mark

The solution set of the equation $\tan ^{-1} x-\cot ^{-1} x=\cos ^{-1}(2-x)$ is

A
$[0,1]$
B
$[-1,1]$
✓
$[1,3]$
D
None of these

Answer

Correct option: C.

$[1,3]$

(c) : Since, $\tan ^{-1} x$ and $\cot ^{-1} x$ exists for all $x \in R$ and $\cos ^{-1}(2-x)$ exists, if $-1 \leq 2-x \leq 1 \Rightarrow 1 \leq x \leq 3$
$\therefore \tan ^{-1} x-\cot ^{-1} x=\cos ^{-1}(2-x)$ is possible only if $1 \leq x \leq 3$
Thus, the solution of given equation is $[1,3]$.

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MCQ 1691 Mark

If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ equals

A
$0$
B
$1$
✓
$6$
D
$12$

Answer

Correct option: C.

$6$

$\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$
$\because \leq \cos ^{-1} x \leq \pi$
$\Rightarrow \cos ^{-1} \alpha=\cos ^{-1} \beta=\cos ^{-1} \gamma=\pi$
$\Rightarrow \alpha=\beta=\gamma=-1$
$\therefore \alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$
$=-1(-1-1)+(-1)(-1-1)+(-1)(-1-1)$
$=2+2+2=6$

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MCQ 1701 Mark

The value of $\sin \left(2 \tan ^{-1}(0.75)\right)$ is equal to

A
$0.75$
B
$1.5$
✓
$0.96$
D
$\sin 1.5$

Answer

Correct option: C.

$0.96$

Let $2 \tan ^{-1}(0.75)=\theta$
$\Rightarrow 0.75=\tan \left(\frac{\theta}{2}\right)$
$\therefore \sin \left(2 \tan ^{-1}(0.75)\right)$
$=\sin \theta=\frac{2 \tan \theta / 2}{1+\tan ^2 \theta / 2}=\frac{2 \times 0.75}{1+(0.75)^2}=\frac{1.50}{1.5625}=0.96$

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MCQ 1711 Mark

The domain of the function defined by $f(x)=\sin ^{-1} \sqrt{x-1}$ is

✓
$[1,2]$
B
$[-1,1]$
C
$[0,1]$
D
None of these

Answer

Correct option: A.

$[1,2]$

(a): We know, $\frac{-\pi}{2} \leq \sin ^{-1} \sqrt{x-1} \leq \frac{\pi}{2}$
$\Rightarrow-1 \leq \sqrt{x-1} \leq 1 \Rightarrow 0 \leq x-1 \leq 1 \Rightarrow 1 \leq x \leq 2$
$\therefore \quad$ Domain of $f(x)$ is $[1,2]$.

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MCQ 1721 Mark

The domain of the function $\cos ^{-1}(2 x-1)$ is

✓
$[0,1]$
B
$[-1,1]$
C
$(-1,1)$
D
$[0, \pi]$

Answer

Correct option: A.

$[0,1]$

(a) : We know, $0 \leq \cos ^{-1}(2 x-1) \leq \pi$
$\Rightarrow-1 \leq 2 x-1 \leq 1 \Rightarrow 0 \leq 2 x \leq 2 \Rightarrow 0 \leq x \leq 1$
$\therefore$ Domain of $\cos ^{-1}(2 x-1)=[0,1]$

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MCQ 1731 Mark

If $\theta=\tan ^{-1} a, \phi=\tan ^{-1} b$ and $a b=-1$, then $|\theta-\phi|$ is equal to

A
0
B
$\frac{\pi}{4}$
✓
$\frac{\pi}{2}$
D
None of these

Answer

Correct option: C.

$\frac{\pi}{2}$

(c) : Given that, $\theta=\tan ^{-1} a, \phi=\tan ^{-1} b$ and $a b=-1$
$\therefore \tan \theta \tan \phi=a b=-1 \Rightarrow \tan \theta=-\cot \phi$
$\Rightarrow \tan \theta=\tan \left(\frac{\pi}{2}+\phi\right) \Rightarrow \theta-\phi=\frac{\pi}{2}$

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MCQ 1741 Mark

The value of $\tan ^{-1}(1)+\tan ^{-1}(0)+\tan ^{-1}(-1)$ is equal to

✓
$\pi$
B
$\frac{5 \pi}{4}$
C
$\frac{\pi}{2}$
D
None of these

Answer

Correct option: A.

$\pi$

(a) : $\tan ^{-1}(1)+\tan ^{-1}(0)+\tan ^{-1}(-1)$
$
=\frac{\pi}{4}+\pi-\frac{\pi}{4}=\pi
$

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MCQ 1751 Mark

If $6 \sin ^{-1}\left(x^2-6 x+8.5\right)=\pi$, then $x$ is equal to

A
$1$
✓
$2$
C
$3$
D
$8$

Answer

Correct option: B.

$2$

We have, $6 \sin ^{-1}\left(x^2-6 x+8.5\right)=\pi$
$\Rightarrow \sin ^{-1}\left(x^2-6 x+8.5\right)=\frac{\pi}{6}$
$\Rightarrow x^2-6 x+8.5=\sin \frac{\pi}{6}=\frac{1}{2}$
$\Rightarrow x^2-6 x+8=0$
$\Rightarrow(x-4)(x-2)=0$
$\Rightarrow x=4 \text { or } x=2$

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MCQ 1761 Mark

The number of triplets $(x, y, z)$ satisfies the equation $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$ is

✓
1
B
2
C
0
D
infinite

Answer

Correct option: A.

(a) : We have, $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$
$\because \quad-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}, \frac{-\pi}{2} \leq \sin ^{-1} y \leq \frac{\pi}{2}$
and $\frac{-\pi}{2} \leq \sin ^{-1} z \leq \frac{\pi}{2}$
$\therefore \quad$ The above condition will satisfy if $\sin ^{-1} x=\sin ^{-1} y=\sin ^{-1} z=\frac{\pi}{2} \Rightarrow x=y=z=1$

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MCQ 1771 Mark

The domain of $\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x$ is

A
$[0,1]$
✓
$[-1,1]$
C
$[0,1)$
D
$(-\infty, \infty)$

Answer

Correct option: B.

$[-1,1]$

(b) : Let $f(x)=\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x$. Then $\operatorname{Dom}(f)=[-1,1] \cap[-1,1] \cap R=[-1,1]$.

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MCQ 1781 Mark

$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=$

A
$\frac{\pi}{2}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{4}$
✓
$\frac{\pi}{6}$

Answer

Correct option: D.

$\frac{\pi}{6}$

Let $\cos ^{-1} \frac{\sqrt{3}}{2}=\theta$
$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$
$\Rightarrow \theta=\frac{\pi}{6} \in[0, \pi]$
$\therefore \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$

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MCQ 1791 Mark

Domain of $\cos ^{-1}[x]$ (where [.] denotes G.I.F.) is

A
$[-1,2]$
✓
$[-1,2)$
C
$(-1,2]$
D
None of these

Answer

Correct option: B.

$[-1,2)$

(b) : Clearly, $-1 \leq[x] \leq 1$
$\Rightarrow-1 \leq x<2 \Rightarrow x \in[-1,2)$

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MCQ 1801 Mark

$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)+4 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
✓
$\frac{4 \pi}{3}$
D
$\frac{3 \pi}{4}$

Answer

Correct option: C.

$\frac{4 \pi}{3}$

(c) : $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)+4 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$
=\frac{\pi}{3}+2 \cdot \frac{\pi}{6}+4 \cdot \frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{4 \pi}{3}
$

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MCQ 1811 Mark

$\cos ^{-1}\left(\frac{-1}{2}\right)+2 \sin ^{-1}\left(\frac{-1}{2}\right)$ is equal to

✓
$\frac{\pi}{3}$
B
$\frac{2 \pi}{3}$
C
$\frac{3 \pi}{4}$
D
$\frac{5 \pi}{8}$

Answer

Correct option: A.

$\frac{\pi}{3}$

(a) : Principal value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ is $\frac{2 \pi}{3}$
and principal value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ is $\left(\frac{-\pi}{6}\right)$.
$\therefore \quad \cos ^{-1}\left(\frac{-1}{2}\right)+2 \sin ^{-1}\left(\frac{-1}{2}\right)$
$=\frac{2 \pi}{3}+\left(2 \times \frac{-\pi}{6}\right)=\frac{2 \pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}$

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MCQ 1821 Mark

Find the principal values of: $\cot ^{-1}(-\sqrt{3})$

✓
$\frac{5 \pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{4}$

Answer

Correct option: A.

$\frac{5 \pi}{6}$

(a) : Let $\cot ^{-1}(-\sqrt{3})=\theta \Rightarrow \cot \theta=-\sqrt{3}=-\cot \frac{\pi}{6}$
$=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6} \Rightarrow \theta=\frac{5 \pi}{6} \in(0, \pi)$
$\therefore$ Principal value of $\cot ^{-1}(-\sqrt{3})$ is $\frac{5 \pi}{6}$.

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MCQ 1831 Mark

Find the principal values of: $\tan ^{-1}(\sqrt{3})$

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{3}$
C
$\frac{2 \pi}{3}$
D
$\frac{5 \pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{3}$

(b) : Let $\tan ^{-1}(\sqrt{3})=\theta \Rightarrow \tan \theta=\sqrt{3}=\tan \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
$\therefore$ Principal value of $\tan ^{-1} \sqrt{3}$ is $\frac{\pi}{3}$.

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MCQ 1841 Mark

Find the principal values of: $\sec ^{-1}(2)$

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{3}$
C
$\frac{2 \pi}{3}$
D
$\frac{5 \pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{3}$

(b) : Let $\sec ^{-1}(2)=\theta \Rightarrow \sec \theta=2=\sec \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\pi}{3} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}$
$\therefore \quad$ Principal value of $\sec ^{-1}(2)$ is $\frac{\pi}{3}$.

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MCQ 1851 Mark

Find the principal values of: $\operatorname{cosec}^{-1}(2)$

✓
$\frac{\pi}{6}$
B
$\frac{2 \pi}{3}$
C
$\frac{5 \pi}{6}$
D
0

Answer

Correct option: A.

$\frac{\pi}{6}$

(a) : Let $\operatorname{cosec}^{-1}(2)=\theta \Rightarrow \operatorname{cosec} \theta=2=\operatorname{cosec} \frac{\pi}{6}$
$\Rightarrow \theta=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
$\therefore \quad$ Principal value of $\operatorname{cosec}^{-1}(2)$ is $\frac{\pi}{6}$.

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MCQ 1861 Mark

Find the principal values of: $\cot ^{-1}(1)$

A
$\frac{\pi}{3}$
✓
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
0

Answer

Correct option: B.

$\frac{\pi}{4}$

(b) : Let $\cot ^{-1}(1)=\theta \Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4} \in(0, \pi)$
$\therefore$ Principal value of $\cot ^{-1}(1)$ is $\frac{\pi}{4}$.

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MCQ 1871 Mark

Find the principal values of: $\cos ^{-1}\left(\frac{1}{2}\right)$

A
$-\frac{\pi}{3}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$\frac{2 \pi}{3}$

Answer

Correct option: B.

$\frac{\pi}{3}$

(b) : Let $\cos ^{-1}\left(\frac{1}{2}\right)=\theta \Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\pi}{3} \in[0, \pi]$
$\therefore$ Principal value of $\cos ^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{3}$.

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MCQ 1881 Mark

Find the principal values of: $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
✓
$\frac{5 \pi}{6}$
D
$\frac{2 \pi}{3}$

Answer

Correct option: C.

$\frac{5 \pi}{6}$

(c) : Let $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\theta \Rightarrow \sec \theta=\frac{-2}{\sqrt{3}}=-\sec \frac{\pi}{6}$
$
\begin{aligned}
& =\sec \left(\pi-\frac{\pi}{6}\right)=\sec \frac{5 \pi}{6} \\
\Rightarrow & \theta=\frac{5 \pi}{6} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}
\end{aligned}
$
$\therefore \quad$ Principal value of $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ is $\frac{5 \pi}{6}$.

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MCQ 1891 Mark

Find the principal values of: $\sin ^{-1}\left(\frac{-1}{2}\right)$

A
$\frac{\pi}{3}$
B
$-\frac{\pi}{3}$
C
$\frac{\pi}{6}$
✓
$-\frac{\pi}{6}$

Answer

Correct option: D.

$-\frac{\pi}{6}$

(d) : Let $\sin ^{-1}\left(\frac{-1}{2}\right)=\theta \Rightarrow \sin \theta=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)$
$\Rightarrow \theta=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Principal value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ is $\left(\frac{-\pi}{6}\right)$.

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MCQ 1901 Mark

Find the principal value of: $\tan ^{-1}(-1)$.

✓
$-\frac{\pi}{4}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{6}$

Answer

Correct option: A.

$-\frac{\pi}{4}$

(a): Let $\tan ^{-1}(-1)=x \Rightarrow-1=\tan x$
We know that the range of principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then, $-1=\tan \left(-\frac{\pi}{4}\right)$, where $-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Hence, the principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$.

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MCQ 1911 Mark

Evaluate : $\sin \left[\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right]$

A
$\sqrt{3} / 2$
B
$1 / 2$
C
$0$
✓
$1$

Answer

Correct option: D.

$1$

$ \sin \left(\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right)$
$=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \frac{\pi}{2}=1$

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MCQ 1921 Mark

The principal solution of $\cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)$ is

A
$\frac{7 \pi}{6}$
✓
$\frac{5 \pi}{6}$
C
$\frac{\pi}{6}$
D
$\frac{11 \pi}{6}$

Answer

Correct option: B.

$\frac{5 \pi}{6}$

$\cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)=\cos ^{-1}\left(\cos \left(\pi+\frac{\pi}{6}\right)\right)$
$=\cos ^{-1}\left(-\cos \left(\frac{\pi}{6}\right)\right)$
$=\cos ^{-1}\left(\cos \left(\pi-\frac{\pi}{6}\right)\right)=\cos ^{-1}\left(\cos \left(\frac{5 \pi}{6}\right)\right)=\frac{5 \pi}{6}$

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MCQ 1931 Mark

Find the value of $\tan ^{-1}\left(2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right)$.

A
$\frac{\pi}{3}$
✓
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{6}$

Answer

Correct option: B.

$\frac{\pi}{4}$

We have,
$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1}\left(\frac{1}{2}\right)\right)\right\}=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}$
$=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\}=\tan ^{-1}\left[2 \times \frac{1}{2}\right]=\tan ^{-1} 1=\frac{\pi}{4}$

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MCQ 1941 Mark

Evaluate: $\cos \left(\frac{\pi}{3}-\cos ^{-1} \frac{1}{2}\right)$

A
0
B
$\frac{1}{2}$
✓
1
D
None of these

Answer

Correct option: C.

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MCQ 1951 Mark

Evaluate : $\cos \left(2 \cos ^{-1}\left(\frac{2}{5}\right)\right)$

A
$\frac{13}{25}$
B
$\frac{17}{25}$
C
$\frac{-13}{25}$
✓
$\frac{-17}{25}$

Answer

Correct option: D.

$\frac{-17}{25}$

(d) : $\cos \left(2 \cos ^{-1}\left(\frac{2}{5}\right)\right)=\cos 2 x$, where $x=\cos ^{-1} \frac{2}{5}$ $=2 \cos ^2 x-1=2\left(\frac{2}{5}\right)^2-1 \quad\left(\because \cos x=\frac{2}{5}\right)$
$=\frac{2 \times 4}{25}-1=\frac{8-25}{25}=-\frac{17}{25}$

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MCQ 1961 Mark

Evaluate : $\operatorname{cosec}^{-1}(2 / \sqrt{3})$

A
$\frac{\pi}{4}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{6}$
D
$\frac{5 \pi}{3}$

Answer

Correct option: B.

$\frac{\pi}{3}$

(b) : $\operatorname{cosec}^{-1}(2 / \sqrt{3})=\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{\pi}{3}\right)\right)=\frac{\pi}{3}$

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MCQ 1971 Mark

Find the principal value of $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

A
$\frac{2 \pi}{3}$
B
$\frac{\pi}{3}$
✓
$\frac{5 \pi}{6}$
D
$\frac{\pi}{4}$

Answer

Correct option: C.

$\frac{5 \pi}{6}$

(c) : Let $x=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$, then $\cos x=\frac{-\sqrt{3}}{2}$
We know that the range of principal value branch of $\cos ^{-1}$ is $[0, \pi]$
$\Rightarrow \cos x=\cos \left(\pi-\frac{\pi}{6}\right)=\frac{-\sqrt{3}}{2} \Rightarrow x=\frac{5 \pi}{6} \in[0, \pi]$
$\therefore \quad$ Principal value is $\frac{5 \pi}{6}$.

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MCQ 1981 Mark

Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

✓
$\frac{\pi}{4}$
B
$\frac{\pi}{6}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{2}$

Answer

Correct option: A.

$\frac{\pi}{4}$

(a) : Let $x=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$, then $\sin x=\frac{1}{\sqrt{2}}$
We know that the range of principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and $\sin x=\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \Rightarrow x=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad$ Principal value is $\frac{\pi}{4}$.

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