M.C.Q (1 Marks)

Question 511 Mark

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

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{2\pi}{3}$
$\frac{5\pi}{6}$

Answer

$\frac{\pi}{3}$

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Question 521 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:

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

Answer

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

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Question 531 Mark

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

$\sin^{-1}\text{⁡x}$
$\cot^{-1}\text{⁡x}$
$\tan^{-1}\text{⁡x}$
$\sec^{-1}\text{⁡x}$

Answer

$\sin^{-1}\text{⁡x}$

Solution:
[-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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Question 541 Mark

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

0
$ π$
$ 8π - 24$
none of these

Answer

8π - 24

Solution:
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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Question 551 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:

$\text{x}+\text{y}=\frac{\pi}{8}$
$\text{x}+\text{y}={2}$
$\text{x}+\text{y}=\frac{\pi}{2}$
$\text{x}+\text{y}=\frac{\pi}{8}$

Answer

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

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

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Question 561 Mark

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

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$-\frac{1}{2}$
$-\frac{\sqrt{3}}{2}$

Answer

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

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Question 571 Mark

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

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

Answer

$\frac{24}{25}$

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Question 581 Mark

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

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

Answer

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

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Question 591 Mark

Consider the following statements:

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

Which of the above statements is/are correct ?

1 only
2 only
Both 1 and 2
Neither 1 nor 2

Answer

2 only

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

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Question 601 Mark

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

$[0,\pi]$
$[\frac{\pi}{4},\frac{3\pi}{4}]$
$(0,\pi)$
$[0,\frac{\pi}{2}]$

Answer

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

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Question 611 Mark

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

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

Answer

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

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Question 621 Mark

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

0
8π - 26
4π + 2
None of these

Answer

8π - 26

Solution:
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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Question 631 Mark

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

$\frac{6}{17}$
$\frac{3}{17}$
$\frac{4}{17}$
$\frac{5}{17}$

Answer

$\frac{6}{17}$

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MCQ 641 Mark

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

A
$160^0$
B
$70^0$
✓
$-20^0$
D
$20^0$

Answer

Correct option: C.

$-20^0$

sinsin of an angle is positive in first and second quadrants.
$ \Rightarrow \sin ^{ -1 }{ (\sin { { 160 }^{ 0 } } } )$
$\Rightarrow(\sin ^{ -1 }{ (\sin { { (180-20) }^{ 0 } } })$
$=20^0$

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Question 651 Mark

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

$\text{e}^{\frac{5\pi}{18}}$
$\text{e}^{\frac{13\pi}{18}}$
$\text{e}^{\frac{-2\pi}{18}}$
$\text{none of these}$

Answer

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

Solution:
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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Question 661 Mark

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

$\text{x}$
$\frac{1}{\text{x}}$
$2\text{x}$
$\frac{2}{\text{x}}$

Answer

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

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Question 671 Mark

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

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

Answer

$\frac{\pi}{4}$

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Question 681 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

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

Answer

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

Solution:
$ \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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Question 691 Mark

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

$\sqrt{\tan\theta}$
$\sqrt{\cot\theta}$
$\tan\theta$
$\cot\theta$

Answer

$\sqrt{\tan\theta}$

Solution:
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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Question 701 Mark

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

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

Answer

$\frac{\pi}{3}$

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

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Question 711 Mark

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

$ \frac { 1 }{ \sqrt { 5 } }$
$ \frac { 1 }{ \sqrt { 7 } }$
$ \frac { 1 }{ \sqrt { 8 } }$
$ \frac { 1 }{ \sqrt { 9 } }$

Answer

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

Solution:
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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Question 721 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:

$\frac{1}{5}$
$\frac{2}{5}$
$0$
$1$

Answer

$\frac{2}{5}$

Solution:
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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Question 731 Mark

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

$ \frac { 1 }{ \sqrt { 17 } }$
$ \frac { 1 }{ \sqrt {- 17 } }$
$ \frac { 1 }{ \sqrt {- 14 } }$
$ -\cos 4$

Answer

$ -\cos 4$

Solution:
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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Question 741 Mark

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

$ \dfrac{\pi}{7}$
$ \dfrac{\pi}{6}$
$ \dfrac{\pi}{4}$
$ \dfrac{\pi}{2}$

Answer

$ \dfrac{\pi}{2}$

Solution:
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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Question 751 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:

$\frac{\pi}{6}$
$\frac{5\pi}{6}$
$\frac{7\pi}{6}$
$1$

Answer

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

Solution:
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 761 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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Question 771 Mark

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

Answer

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Question 781 Mark

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

Answer

Solution:
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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Question 791 Mark

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

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

Answer

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Question 801 Mark

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

π
2π
$ \cfrac{3\pi}{2}$
None of these

Answer

Solution:
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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Question 811 Mark

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

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

Answer

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

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Question 821 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:

1
2
3
Infinite

Answer

Solution:
$ 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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Question 831 Mark

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

$\frac{\pi}{3 }$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
None of these.

Answer

$\frac{\pi}{6}$

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Question 841 Mark

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

No solution.
Only one solution.
Two solutions.
Three solutions.

Answer

No solution.

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Question 851 Mark

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

$\sin^2\alpha$
$\cos^2\alpha$
$\tan^2\alpha$
$\cot^2\alpha$

Answer

$\sin^2\alpha$

Solution:
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
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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Question 861 Mark

$\sin−10 $ is equal to:

$0$
$ \dfrac{\pi }{6}$
$ \dfrac{\pi}{2}$
$ \dfrac{\pi}{3}$

Answer

Solution:
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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Question 871 Mark

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

$ \alpha = 0, \beta = \pi(2)$
$ \alpha = 0, \beta = 2\pi$
$ \alpha = 0, \beta = \pi$
$ \alpha = 0, \beta = 5\pi$

Answer

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

Solution:
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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Question 881 Mark

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

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

Answer

$\frac{\pi}{3}$

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Question 891 Mark

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

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

Answer

$\frac{\sqrt3}{2}$

Solution:
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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Question 901 Mark

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

$-1$
$\frac{2}{5}$
$\frac{1}{3}$
$\frac{1}{5}$

Answer

$\frac{1}{5}$

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Question 911 Mark

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

$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{5\pi}{2}$
$\frac{\pi}{6}$

Answer

$\frac{\pi}{4}$

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

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Question 921 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:

0
1
2
infinite

Answer

Solution:
$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$
$\Rightarrow\sqrt{2\cos^2\text{x}}=\sqrt2(-\pi-\text{x})$
$\Rightarrow|\cos\text{x}|=\text{x}$
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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Question 931 Mark

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

$\frac{\pi}{6}$
$\frac{5\pi}{6}$
$\frac{7\pi}{6}$
$1$

Answer

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

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Question 941 Mark

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

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

Answer

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

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Question 951 Mark

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

0
$\frac{1}{2}$
-1
None of these

Answer

None of these

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

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Question 961 Mark

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

$\text{x}$
$\sqrt{1-\text{x}^2}$
$\frac{1}{\text{x}}$
$\text{none of these}$

Answer

$\text{x}$

Solution:
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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Question 971 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:

5
$ \frac{\pi}{2}$
0
π

Answer

$ \frac{\pi}{2}$

Solution:
$\text{x}\in \left ( \frac{3\pi}{2}, 2\pi \right )$
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}$

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Question 981 Mark

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