M.C.Q (1 Marks)

Question 1011 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:

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

Answer

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

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Question 1021 Mark

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

Answer

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Question 1031 Mark

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

$ \frac {\sqrt {2}}{4}$
$ \frac {\sqrt {2}}{6}$
$ \frac {\sqrt {6}}{4}$
$ \frac {\sqrt {2}}{3}$

Answer

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

Solution:
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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Question 1041 Mark

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

0
−1
2
1

Answer

Solution:

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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Question 1051 Mark

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

all x
x > 0
$\text{x }\epsilon[-1,1]$
$\frac{1}{\sqrt{2}}\leq\text{x}\leq1$

Answer

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

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Question 1061 Mark

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

$0$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{2}$

Answer

$\frac{\pi}{4}$

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Question 1071 Mark

If $\text{x }\epsilon\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),$ 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:

$\frac{\text{x}}{2}$
2x
3x
x

Answer

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Question 1081 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:

$\frac{\pi}{5}$
$\frac{2\pi}{5}$
$\frac{3\pi}{5}$
$\pi$

Answer

$\frac{\pi}{5}$

Solution:
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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Question 1091 Mark

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

0.81
0.56
0.48
0.28

Answer

0.56

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Question 1101 Mark

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

$\frac{2\pi}{3}$
$\frac{3\pi}{4}$
$\frac{\pi}{2}$
$6\pi$

Answer

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

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Question 1111 Mark

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

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

Answer

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

Solution:
$\sin^{-1}(\text{x})$
$-\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big),$ 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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Question 1121 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?

0
1
2
Infinitely many

Answer

Solution:
$\tan ^{-1} \frac{\text{x}^4+1-\text{x}^2}{\text{x}^3-\text{x}^5-\text{x}}$
$$$= ^{-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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Question 1131 Mark

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

$\pm\frac{\pi}{3}$
$\pm\frac{\pi}{4}$
$\pm\frac{\pi}{6}$
$\text{none of these}$

Answer

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

Solution:
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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Question 1141 Mark

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

$\frac{3\pi}{5}$
$\frac{-3\pi}{5}$
$\frac{\pi}{10}$
$\frac{-\pi}{10}$

Answer

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

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Question 1151 Mark

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

$5$
$\frac{1}{5}$
$\frac{5}{14}$
$\frac{14}{5}$

Answer

$\frac{1}{5}$

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Question 1161 Mark

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

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

Answer

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

Solution:
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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Question 1171 Mark

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

$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{2}$

Answer

$\frac{\pi}{4}$

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Question 1181 Mark

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

$\frac{25}{24}$
$\frac{25}{7}$
$\frac{24}{25}$
$\frac{7}{24}$

Answer

$\frac{7}{24}$

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

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MCQ 1191 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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Question 1201 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:

$\frac{3\pi}{5}$
$\frac{-7\pi}{5}$
$\frac{\pi}{10}$
$\frac{-\pi}{10}$

Answer

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

Solution:
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 1211 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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Question 1221 Mark

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

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

Answer

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

Solution:
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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Question 1231 Mark

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

$\frac{19}{8}$
$\frac{8}{19}$
$\frac{19}{2}$
$\frac{3}{4}$

Answer

$\frac{19}{8}$

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

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MCQ 1241 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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Question 1251 Mark

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

$ \frac{\pi}{2}$
$ \frac{\pi}{3}$
$ \frac{\pi}{4}$
$ \frac{\pi}{6}$

Answer

$ \frac{\pi}{2}$

Solution:
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 }$

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Question 1261 Mark

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

$[3,5]$
$[-1,1]$
$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
$\Big[-\sqrt5,-\sqrt3\Big]\cap\Big[\sqrt3,\sqrt5\Big]$

Answer

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

Solution:
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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Question 1271 Mark

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

$4\alpha=3\beta$
$3\alpha=4\beta$
$\alpha-\beta=\frac{7\pi}{12}$
$\text{none of these}$

Answer

$4\alpha=3\beta$

Solution:
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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Question 1281 Mark

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

$\frac{1}{\sqrt2}$
$\frac{1}{\sqrt3}$
$\frac{1}{2\sqrt2}$
$\frac{1}{3\sqrt3}$

Answer

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

Solution:
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
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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Question 1291 Mark

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

$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{3\pi}{4}$

Answer

$\frac{\pi}{4}$

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Question 1301 Mark

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

$\frac{1}{2}$
$\frac{\sqrt3}{2}$
$-\frac{1}{2}$
$\text{none of these}$

Answer

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

Solution:
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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Question 1311 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:

x - y - xy = 1
x - y + xy = 1
x + y - xy = 1
x + y + xy = 1

Answer

x + y + xy = 1

Solution:
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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Question 1321 Mark

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

$0$
$1$
$-1$
$\frac{1}{2}$

Answer

Solution:
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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Question 1331 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]
[-1, 1]
[0, 1]
none of these.

Answer

[1, 2]

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

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Question 1341 Mark

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

7
6
5
none of these

Answer

Solution:
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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Question 1351 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
x = 2, y = 1
x = 3, y = 2
x = -2, y = -1

Answer

x = 1, y = 2

Solution:
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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Question 1361 Mark

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

$ \cos ^{-1}\text{x}$
$ \sec ^{-1}\text{x}$
$ \text{cosec} ^{-1}\text{x}$
$ \tan ^{-1}\text{x}$

Answer

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

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Question 1371 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:

$4\tan^{-1}\text{x}$
$0$
$\frac{\pi}{2}$
$\pi$

Answer

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

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

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Question 1381 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:

$0$
$1$
$2$
$\infty$

Answer

Solution:
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 1391 Mark

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

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

Answer

Correct option: C.

$-1\leq\text{x}<\frac{1}{\sqrt2}$

The correct option is $C$
$
-1 \leq x < \frac{1}{\sqrt{2}}
$
Explanation for the correct options:
Step $1$.
Find the intervals in which $x$ lies:
We have given $\cos ^{-1} x > \sin ^{-1} x$, and we know that,
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x$
$\text { But } \frac{\pi}{2}-\sin ^{-1} x > \sin ^{-1} x$
$\Rightarrow \frac{\pi}{2} > 2 \sin ^{-1} x$
$\Rightarrow \frac{\pi}{4} > \sin ^{-1} x \ldots \ldots \ldots(1)$
Also $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \ldots \ldots(2)$

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Question 1401 Mark

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

[1, 2]
[-1, 1]
[0, 1]
None of these.

Answer

[1, 2]

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Question 1411 Mark

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

$\frac{6}{25}$
$\frac{24}{25}$
$\frac{4}{5}$
$-\frac{24}{25}$

Answer

$-\frac{24}{25}$

Solution:
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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Question 1421 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}} :$

$ \frac{\pi}{2}$
$ \frac{\pi}{3}$
$ \frac{\pi}{6}$
$ \frac{\pi}{8}$

Answer

$ \frac{\pi}{6}$

Solution:
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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Question 1431 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:

$0$
$\frac{\text{a}}{2}$
$\text{a}$
$\frac{2\text{a}}{1-\text{a}^2}$

Answer

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

Solution:
$\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
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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Question 1441 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:

$\frac{\pi}{2}$
$\frac{3\pi}{2}$
$\frac{5\pi}{2}$
$\frac{7\pi}{2}$

Answer

$\frac{\pi}{2}$

Solution:
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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Question 1451 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:

$0$
$\frac{\text{a}}{2}$
$\text{a}$
$\frac{2\text{a}}{1-\text{a}^2}$

Answer

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

Solution:
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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Question 1461 Mark

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

Answer

Solution:
${ \sin }^{ -1 }\frac { x }{ 5 } +{ \text{cosec} }^{ -1 }\frac { 5 }{ 4 } =\frac{ \pi }{ 2 }$
$ \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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Question 1471 Mark

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

$ \frac { \pi }{ 3 }$
$ \frac { \pi }{ 4 }$
$ \frac { \pi }{ 6 }$
$ \frac { \pi }{ 9 }$

Answer

$ \frac { \pi }{ 4 }$

Solution:
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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Question 1481 Mark

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

$\frac{1}{6}$
$1$
$(\frac{1}{6},-1)$
None of these.

Answer

$\frac{1}{6}$

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Question 1491 Mark

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

$\frac{33}{5}$
$-\frac{\pi}{10}$
$\frac{\pi}{10}$
$\frac{7\pi}{5}$

Answer

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

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

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Question 1501 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 $\alpha-\beta=$

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$-\frac{\pi}{3}$

Answer

$\frac{\pi}{6}$

Solution:
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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MATHS M.C.Q (1 Marks)