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Maths M.C.Q (1 Marks)

INVERSE TRIGNOMETRIC FUNCTIONS · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 152 questions.

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M.C.Q (1 Marks)

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152 questions · auto-graded multiple-choice test.

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.
$\frac{-2\pi}{5}$
$\tan^{-1}\Big(\tan\Big(\frac{3\pi}{5}\Big)\Big)$

Let $\text{y}=\tan^{-1}\Big(\tan\Big(\frac{3\pi}{5}\Big)\Big)$

$\Rightarrow\tan\text{y}=\tan\Big(​​\frac{3\pi}{5}\Big)$

$\Rightarrow\tan\text{y}=\tan(108^\circ)$

We know that the range of principal value of $\tan^{-1}$ is $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$

Hence y = 108º not possible.

Now, $\tan\text{y}=\tan(108^\circ)$

$\Rightarrow\tan\text{y}=\tan(180^\circ-72^\circ)$

$\Rightarrow\tan\text{y}=-\tan(72^\circ)$ $(\text{as}\tan(180^\circ-\theta)=-\tan\theta)$

$\Rightarrow\tan\text{y}=\tan(72^\circ)$ $(\text{as}\tan(-\theta)=-\tan\theta)$

$\Rightarrow\tan\text{y}=\tan\Big(-72^\circ\times\frac{\pi}{180}\Big)$

$\Rightarrow\tan\text{y}=\tan\Big(\frac{-2\pi}{5}\Big)$

$\Rightarrow\text{y}=\frac{-2\pi}{5}$
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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.
$-\frac{24}{25}$
Let $\cos^{-1}\Big(-\frac{3}{5}\Big)=\text{x},0\leq\text{x}\leq\pi$

Then, $\cos\text{x}=-\frac{3}{5}$
$\therefore\ \sin\text{x}=\sqrt{1-\cos^2\text{x}}=\sqrt{1-\Big(-\frac{3}{5}\Big)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now,

$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}=\sin(2\text{x})$

$=2\sin\text{x}\cos\text{x}$

$=2\times\frac{4}{5}\times\frac{-3}{5}$

$=-\frac{24}{25}$
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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.
$\frac{1}{\sqrt2}<\text{x}\leq1$
$\cos^{-1}\text{x}>\sin^{-1}\text{x}$

$\Rightarrow\cos^{-1}\text{x}>\frac{\pi}{2}-\cos^{-1}\text{x}$

$\Rightarrow2\cos^{-1}\text{x}>\frac{\pi}{2}$

$\Rightarrow\cos^{-1}\text{x}>\frac{\pi}{4}$

$\Rightarrow\text{x}>\cos\frac{\pi}{4}$

$\Rightarrow\text{x}>\frac{1}{\sqrt2}$

We know that the maximum value of cosine function is 1.

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

Hence, the correct ans is option (a).
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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: Hint: $\bigg[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\bigg]$
  • A
    $2+\sqrt{5}$
  • ✓
    $\sqrt{5}-2$
  • C
    $\frac{\sqrt{5}+2}{2}$
  • D
    $5+\sqrt{2}$
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} $

⇒ 3x + 1 = A (x + 3) + B(x - 1)

Substitute x = 1 both sides

⇒ 3(1) + 1 = A(1 + 3) + 0 ⇒ A = 1

Substitute x = - 3x both sides

⇒ 3( -3) + 1 = 0 + B(-3 -1)

⇒ -8 - 4B ⇒ 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
    $\text{none of these}$
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 $\ce{cosec}^{-1}\ x?$
  • 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 $\ce{cosec}^{-1}\ x$ 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.
$0$
View full question & answer→
MCQ 291 Mark
Choose the correct answer from the given four options.The domain of the function $\cos^{-1}(2x - 1)$ is:
  • ✓
    $[0,1]$
  • B
    $[-1,1]$
  • C
    $(-1,1)$
  • D
    $[0,\pi]$
Answer
Correct option: A.
$[0,1]$
We have, $\cos^{-1}(2x - 1)$
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, $\sin^{-1}\text{x}=θ.$

Then, $\cos 2\theta =\frac{1}{9}​$

$ 1-2\sin ^{2}\theta =\frac{1}{9}$

or $1-2\text{x}^{2}=\frac{1}{9}$

$\text{x}^{2}=\frac{4}{9}\text{x}$

$ ∴ \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.
$\pm\sqrt{\frac{5}{2}}$
View full question & answer→
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
    π
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
    $\text{none of these}$
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.
$0.96$
View full question & answer→
MCQ 391 Mark
Domain of $ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +\text{c}o\sec ^{ -1 }{ \text{x}}$ is:
  • A
    [-1, 1]
  • B
    R
  • C
    (-∞, -1] ∪ [1, ∞)
  • ✓
    {-1, 1}
Answer
Correct option: D.
{-1, 1}
$ \text{f}(\text{x})=\cot ^{ -1 }{ \text{x} } +\cos ^{ -1 }{ \text{x} } +co\sec ^{ -1 }{ \text{x}}$

Domain of $\cot^{−1}\text{x}=(−∞,∞)$

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 $\text{f}_1​(\text{x}) ∩$

domain of $\text{f}_2(\text{x}) ∩$

domain of $\text{f}_\text{n}(\text{x})$
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MCQ 401 Mark
The value of $\cot^{-1}9+\text{cosec}^{-1}\Big(\frac{\sqrt{41}}{4}\Big)$ 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 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 } \text{if} \cos-1 \text{x} + \cos-1 \text{y} + \cos-1 \text{z} = 3π?$
  • 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} = 3π$

This means $ \cos-1 \text{x} = π, \cos-1 \text{y} = π$ and $ \cos-1 \text{z} = π$

This will be only possible when it is in maxima.

As, $\cos-1 \text{x} = π$ so,$ \text{x} = \cos-1 π = -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 (-∞, ∞).

The domain for $\tan^{-1}\text{⁡x}$ is (-∞, ∞).

The domain for $\sec^{-1}\text{⁡x}$ is (-∞, -1) ∪ (1, ∞).
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MCQ 471 Mark
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 12)$ is:
  • A
    0
  • B
    $ π$
  • ✓
    $ 8π - 24$
  • D
    none of these
Answer
Correct option: C.
$ 8π - 24$
12 rad lies in 4th quadrant

$ \frac{7\pi}{2}<12<4\pi$

Let θ be an acute angle such that

$ 12+\theta=4\pi$

$∴12=4π−θ or \theta=4\pi-12θ=4π−12$

$ \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(4π−24)$

$ =8π−24​$

$ ∴\cos−1(\cos12)−\sin−1(\sin12)=8π−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
    $ 2π​$
  • C
    $ 3π​$
  • 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
    -2π - 6
  • B
    2π + 6
  • ✓
    either -2π + 6 or 2π + 6
  • D
    2π - 6
Answer
Correct option: C.
either -2π + 6 or 2π + 6
We know that $\sin(\text{x}) = \sin(2\text{A} * π + \text{x})$ where A can be positive or negative integer.

If A is -1, then $ \sin(6) = \sin(-2π + 6);$

If A is 1, then $ \sin(6) = \sin(2π + 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).
View full question & answer→
MCQ 591 Mark
$ \tan^{−1}\sqrt{3}+\sec−12–\cos−^{1}1$ is equal to ________.
  • A
    0
  • ✓
    $ \frac{2}{π3}$
  • C
    $ \frac{\pi}{3}$
  • D
    $ \frac{\pi}{4}$
Answer
Correct option: B.
$ \frac{2}{π3}$
$\tan^{-1}\sqrt{3}=\frac{\pi}{3},\sec^{-1}2,\cos^{-1}1=0$

$ ∴\tan^{−1}\sqrt{13}+\sec^{−1}2−\cos^{−1}1=\frac{π}{3}+\frac{π}{3}−0$

$ =\frac{2π}{3}$
View full question & answer→
MCQ 601 Mark
Consider $ \text{x} = 4\tan^{-1}\left (\frac {1}{5}\right ), \text{y} = \tan^{-1} \left (\frac {1}{70}\right )\text{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)\\$
View full question & answer→
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.
$\tan^{-1}\Big(\frac{3\text{a}-\text{a}^3}{1-3\text{a}^2}\Big)$
View full question & answer→
MCQ 621 Mark
Consider the following statements:
  1. $\tan^{-1} 1+ \tan^{-1} (0.5) = \dfrac {\pi}2$
  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 $\tan^{-1} { \text{x} } + \cot^{-1} { \text{x} } =\frac { \pi }{ 2 } \text{x} \in \text{R and}\sin^{-1}{\text{x}} + \cos^{-1}{\text{x}} =\frac{\pi}{2}$,
​and $ \sin-1\frac{1}{3}+\cos-1{\frac{1}{3}} =\cfrac{\pi}{2}$
Hence, only second statement is correct.
View full question & answer→
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.
$[\frac{\pi}{4},\frac{3\pi}{4}]$
View full question & answer→
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 $[0,\pi].$
View full question & answer→
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.
$\frac{2\pi}{3}$
View full question & answer→
MCQ 661 Mark
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$ is
  • A
    0
  • ✓
    8π - 26
  • C
    4π + 2
  • D
    None of these
Answer
Correct option: B.
8π - 26
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$

$ =\cos^{−1}(\cos(4π−12))−\sin^{−1}(−\sin(4π−14))$

$ =4π − 12 − 14 + 4π = 8π − 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}$
View full question & answer→
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 }$
$⇒\sin^{−1}\text{x}+\sin^{−1}\text{y}=\sin ^{ -1 } \Big(\text{x}\sqrt { 1-{ \text{y} }^{ 2 } } +\text{y}\sqrt { 1-\text{x}^{ 2 } }\Big)$
$ \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 }$
View full question & answer→
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,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$

$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$

$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$

$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$

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

$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$

$=\frac{\pi}{4}$
View full question & answer→
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.
$\frac{5\pi}{6}$
View full question & answer→
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.
$\sqrt{\frac{3}{76}}$
View full question & answer→
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
$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$

$=\tan^{-1}\Bigg(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{2}{11}\times\frac{1}{11}}\Bigg)$

$=\tan^{-1}\Bigg(\frac{\frac{3}{11}}{1-\frac{2}{121}}\Bigg)$

$=\tan^{-1}\Big(\frac{33}{119}\Big)$
View full question & answer→
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
    $\text{none of these}$
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
    π
Answer
Correct option: B.
$ \frac{\pi}{2}$
$\text{x}\in \left ( \frac{3\pi}{2}, 2\pi \right )$

Now, $ \cos^{-1}(\cos \,\text{ x})=2π−\text{x}$

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

$ ∴\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{π​}{2}$
View full question & answer→
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
    ∞
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 2π only if $ \sin ^{-1}\text{x}=\sin ^{-1}\text{y}=\frac{\pi }{2} \text{and } \cos^{-1}\text{z}=\pi$

This ispossible only if x = y = 1and z = -1

Hence there is only one solutionf $ (1, 1, -1)= 2π.$
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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.
$\frac{\sqrt{3}}{2}$
View full question & answer→
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.
$\frac{6}{17}$
View full question & answer→
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.
$\frac{1}{\sqrt{2}}\leq\text{x}\leq1$
View full question & answer→
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.
$\frac{\pi}{4}$
View full question & answer→
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.
View full question & answer→
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.
$\tan^{-1}(15)$
View full question & answer→
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.
$2$
View full question & answer→
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,

$3\sin^{-1}\Big(\frac{\tan2\text{y}}{1+\tan^2\text{y}}\Big)-4\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)+2\tan^{-1}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)=\frac{\pi}{3}$

$\Rightarrow3\sin^{-1}(\sin2\text{y})-4\cos^{-1}(\cos2\text{y})+2\tan^{-1}(\tan2\text{y})=\frac{\pi}{3}$

$\Big[\because\ \sin2\text{y}=\Big(\frac{2\tan\text{y}}{1+\tan^2\text{y}}\Big),\cos2\text{y}=\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)\text{ and }\tan2\text{y}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)\Big]$

$\Rightarrow3\times2\text{y}-4\times2\text{y}+2\times2\text{y}=\frac{\pi}{3}$

$\Rightarrow6\text{y}-8\text{y}+4\text{y}=\frac{\pi}{3}$

$\Rightarrow2\text{y}=\frac{\pi}{3}$

$\Rightarrow\text{y}=\frac{\pi}{6}$

$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$ $\big[\because\ \tan^{-1}\text{x}=\text{y}\big]$

$\Rightarrow\text{x}=\tan\frac{\pi}{6}$

$\Rightarrow\text{x}=\frac{1}{\sqrt3}$
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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, $\cos^{-1}\big(\text{x}^2-4\big)=\text{y}$
$\Rightarrow\cos\text{y}=\text{x}^2-4$

$\Rightarrow-1\leq\text{x}^2-4\leq1$

$\Rightarrow3\leq\text{x}^2\leq5$

$\Rightarrow\pm\sqrt3\leq\text{x}\pm\sqrt5$

$\text{x}\in\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
View full question & answer→
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.
$x$
View full question & answer→
MCQ 881 Mark
$ \sin^{-1}\text{⁡x}+\cos^{1}\text{⁡x}= $
  • ✓
    $ \frac{π}{2}$
  • B
    π
  • C
    π3
  • D
    2π
Answer
Correct option: A.
$ \frac{π}{2}$
$ \sin-1\text{⁡x}+\cos-1\text{⁡x}=π2; \text{x} ∈ [-1,1] $
View full question & answer→
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})\text{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.
$[1, 2]$
View full question & answer→
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, $\text{x}=\cos^{-1}(\cos 4)$
$ = 2π - 4$

and $\text{y}=\sin^{-1}(\sin 3)\text{y}$

$ π - 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} } =π$ 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
    $\text{none of these}$
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
    $\text{none of these}$
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 }π$
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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, $ θ = \cos-1(\text{-x})$
So, $ 0 ≤ θ ≤ π$

$ ⇒ -\text{x} = \cosθ$

$ ⇒ \text{x} = -\cosθ$

$ ⇒ \text{x} = \cos-θ$

Also, $ -π ≤ -θ ≤ 0$

So, $ 0 ≤ π -θ ≤ π$

$ ⇒ -θ = \cos^{-1}(\text{x})$

$ ⇒ θ = \cos^{-1}(\text{x})$

So, $\cos-1(\text{x}) = π – θ$

$ θ = π – \cos-1(\text{x})$

$ ⇒ \cos^{-1}(-\text{x}) = π – \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
    $\text{none of these}$
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)$
$=\tan^{-1}\pi+\cot^{-1}\pi+\tan^{-1}\sqrt 3-\sec^{-1}(-2)$

$ [∵\tan^{−1}\frac{1}{\text{y}​}=\cot−1\text{y}]$

$ =\cfrac{\pi}{2}+\tan^{-1}\sqrt 3-\sec^{-1}(-2)\Big[∵\tan^{−1}x+\cot^{−1}\text{x}=\frac{π}{2}\Big]=\frac{π}{2}+\frac{π}{3}-\frac{2π}{3}$
$ =[∵\tan\frac{π}{3}​=\sqrt{3}​;\sec\frac{2π}{3}​=−2]$

$ =\frac{π​}{2}−\frac{π​}{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
    $\text{none of these}$
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]
$\text{f}(\text{x})=\sin^{-1}\sqrt{\text{x}-1}$

$\Rightarrow\ 0\leq\text{x}-1\leq1$ $[\because\ \sqrt{\text{x}-1}\geq0\ \text{and}\ -1\leq\sqrt{\text{x}-1}\leq1]$

$\Rightarrow\ 1\leq\text{x}\leq2$

$\therefore\ \text{x}\in[1,2]$
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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
$\sin^{−1}\text{x}+\sin^{−1}\frac{1}{\text{x}}+\cos^{−1}\text{x}+\cos^{−1}\frac{1}{\text{x}}=$
  • ✓
    π
  • B
    2π
  • C
    $ \cfrac{3\pi}{2}$
  • D
    None of these
Answer
Correct option: A.
π
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

$\sin^{-1}\sqrt{\text{x} ^2-\text{x}+1}-1≤\sqrt{\text{x}^2-\text{x}}+1≤1$

$ 0 ≤\text{x}^2-\text{x}+1≤1$

$\text{x}∈[0,1]$

For $\cos^{-1}\sqrt{\text{x}^2-\text{x}0}≤\text{x}^2-\text{x}≤1$

$ ⇒\text{x}^2−\text{x}≥0$

$ ⇒\text{x}−\text{x}≥0$

$\text{x}∈[−∞,0]∪[1,∞]$ 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 $\text{x}\in\Big[\frac{\pi}{2},\pi\Big],\ \cos\text{x}\leq0$

$\therefore$ cosx = 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\arcsin { \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,

$\text{tan}^1\text{a}+\text{cot}^{-1}\text{a}=\frac{\pi}{2}$

Therefore,

$\text{cot}(\text{tan}^{−1}\text{a}+\text{cot}^{−1}a)=\text{cot}\frac{\pi}{2}=0$
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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) },[∵\sin(2π​−θ)=\cosθ]$

$ =\sin ^{ -1 }{ \left( \sin \left(-\frac{\pi}{10 }\right) \right) } =−\frac{\pi}{10}$

Note $ \sin^{-1}(\sin \theta)=θ \text{ if} -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$
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