M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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50 questions · 2 auto-graded MCQ + 48 self-marked written.

Question 11 Mark

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

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

Answer

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

Solution:
$\tan^{-1}\Big(\tan\Big(\frac{3\pi}{5}\Big)\Big)$
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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Question 21 Mark

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

$36$
$36-36\cos\theta$
$18-18\cos\theta$
$18+18\cos\theta$

Answer

$18-18\cos\theta$

Solution:
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
$\Rightarrow\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2}$
$\Rightarrow\cos^{-1}\Bigg(\frac{\text{x}}{3}\times\frac{\text{y}}{2}-\sqrt{1-\Big(\frac{\text{x}}{3}\Big)^2}\sqrt{1-\Big(\frac{\text{y}}{2}\Big)^2}\Bigg)=\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}}{6}-\sqrt{1-\Big(\frac{\text{x}^2}{9}\Big)}\sqrt{1-\Big(\frac{\text{y}^2}{4}\Big)}=\cos\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}-6\cos\frac{\theta}{2}}{6}=\frac{\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}}{6}$
$\Rightarrow\text{xy}-6\cos\frac{\theta}{2}=\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}$
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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Question 31 Mark

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

Nique solution.
No solution.
Infinitely many solution.
None of these.

Answer

Nique solution.

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Question 41 Mark

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

$\frac{\pi}{2}$
$\frac{5\pi}{3}$
$\frac{10\pi}{3}$
$0$

Answer

0

Solution:
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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Question 51 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:

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

Answer

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

Solution:
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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Question 61 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:

$\sqrt{\text{ab}}$
$\sqrt{2\text{ab}}$
$2\text{ab}$
$\text{ab}$

Answer

$\sqrt{\text{ab}}$

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Question 71 Mark

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

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

Answer

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

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Question 81 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:

Exactly one root.
Almost one root.
At least one root.
No root.

Answer

At least one root.

Solution:
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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Question 91 Mark

The given graph is for which equation?

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

Answer

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

Solution:
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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Question 101 Mark

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

0.75
1.5
0.96
$\sin1.5$

Answer

0.96

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Question 111 Mark

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

[-1, 1]
R
(-∞, -1] ∪ [1, ∞)
{-1, 1}

Answer

{-1, 1}

Solution:
$ \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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Question 121 Mark

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

0.75
1.5
0.96
sin1.5

Answer

0.96

Solution:
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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Question 131 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:

$\frac{1}{\sqrt{2}}$
$-\frac{1}{\sqrt2}$
$\pm\sqrt{\frac{5}{2}}$
$\pm\frac{1}{2}$

Answer

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

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Question 141 Mark

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

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

Answer

$\frac{\pi}{4}$

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Question 151 Mark

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

-2
4
-3
5

Answer

4

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Question 161 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?$

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

Answer

-3

Solution:
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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Question 171 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:

0
-2
1
2

Answer

2

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

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Question 181 Mark

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

2
3
1
none of these

Answer

2

Solution:
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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Question 191 Mark

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

$\cot^{-1}\text{x}$
$\cot^{-1}\text{x}$
$\tan^{-1}\text{x}$
$\text{none of these}$

Answer

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

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

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Question 201 Mark

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

7
6
5
None of these.

Answer

7

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Question 211 Mark

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

$\frac{\sqrt{29}}{3}$
$\frac{29}{3}$
$\frac{\sqrt3}{29}$
$\frac{3}{29}$

Answer

$\frac{3}{29}$

Solution:
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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Question 221 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
$ 2π$
$ 3π$
none of these

Answer

0

Solution:
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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Question 231 Mark

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

0
$ \frac{2}{π3}$
$ \frac{\pi}{3}$
$ \frac{\pi}{4}$

Answer

$ \frac{2}{π3}$

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

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Question 241 Mark

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

$ \frac {\sqrt4}{8}$
$ \frac {\sqrt4}{3}$
$ \frac {\sqrt5}{4}$
$ \frac {\sqrt5}{3}$

Answer

$ \frac {\sqrt5}{3}$

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

$ = \cos \left( \cos^{-1} \left( \frac {\sqrt5}{3} \right) \right)$
$ = \frac {\sqrt5}{3}$

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Question 251 Mark

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

1
0
2
∞

Answer

1

Solution:
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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Question 261 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

$\frac{1}{5}$

Solution:
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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Question 271 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:

$\frac{1}{\sqrt3}$
$-\frac{1}{\sqrt3}$
$\sqrt3$
$-\frac{\sqrt3}{4}$

Answer

$\frac{1}{\sqrt3}$

Solution:
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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Question 281 Mark

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

$ \frac{π}{2}$
π
π3
2π

Answer

$ \frac{π}{2}$

solution:
$ \sin-1\text{⁡x}+\cos-1\text{⁡x}=π2; \text{x} ∈ [-1,1] $

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Question 291 Mark

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

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

Answer

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

Solution:
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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Question 301 Mark

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

x - y = 1
x + y + 1 = 0
x + 2y = 2
y + x = 0

Answer

x + 2y = 2

Solution:
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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Question 311 Mark

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

$\frac { 4 }{ \sqrt { 2 } }$
$ -\frac { 1 }{ \sqrt { 2 } }$
$\frac { 1 }{ \sqrt { 2 } }$
$\frac { 1 }{ \sqrt { 4 } }$

Answer

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

Solution:
$ \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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Question 321 Mark

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

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

Answer

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

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Question 331 Mark

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

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

Answer

$-\frac{24}{25}$

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Question 341 Mark

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

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

Answer

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

Solution:
$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$

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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Question 351 Mark

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

$2+\sqrt{5}$
$\sqrt{5}-2$
$\frac{\sqrt{5}+2}{2}$
$5+\sqrt{2}$

Answer

$\sqrt{5}-2$

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Question 361 Mark

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

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

Answer

$0$

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Question 371 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:

2
3
4
5

Answer

5

Solution:
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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Question 381 Mark

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

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

Answer

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

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MCQ 391 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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Question 401 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}$
2
3
π

Answer

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

Solution:
$ \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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Question 411 Mark

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

$\frac{\pi}{2}$
$-\frac{\pi}{2}$
$-\pi$
$\text{none of these}$

Answer

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

Solution:
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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Question 421 Mark

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

11
15
17
21

Answer

15

Solution:
$ \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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Question 431 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:

$2+\sqrt{5}$
$\sqrt{5}-2$
$\frac{\sqrt{5}+2}{2}$
$5+\sqrt{2}$

Hint: $\bigg[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\bigg]$

Answer

$\sqrt{5}-2$

Solution:
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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Question 441 Mark

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

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

Answer

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

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Question 451 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:

$36$
$-36\sin^2\theta$
$36\sin^2\theta$
$-36\cos^2\theta$

Answer

$36\sin^{2}\theta$

Solution:
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 461 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 $\alpha=\beta=\gamma=\cos\pi\ \text{or }-1$
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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Question 471 Mark

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

$\sqrt2-1$
$1+\sqrt2$
$\frac{\pi}{4}$
$\frac{3\pi}{4}$

Answer

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

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Question 481 Mark

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

-2π - 6
2π + 6
either -2π + 6 or 2π + 6
2π - 6

Answer

either -2π + 6 or 2π + 6

Solution:
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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Question 491 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?

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

Answer

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

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

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Question 501 Mark

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

$ \frac {- \pi }{ 1 }$
$ \frac {- \pi }{ 7 }$
$ \frac { \pi }{ 10 }$
$ \frac {- \pi }{ 10 }$

Answer

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

Solution:
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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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip