Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$\dfrac {\pi}{6}$$

B
$$\dfrac {\pi}{4}$$

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{2}$$

Solution

$$\tan^{-1}\left(2\sin\left(\sec^{-1}(2)\right)\right)$$
Let $$x=\sec^{-1}(2)$$
$$\tan^{-1}\left(2\sin x\right)$$ ----- ( 1 )
$$x=\sec^{-1}(2)$$
$$\Rightarrow$$ $$\sec x=2$$
$$\Rightarrow$$ $$\dfrac{1}{\cos x}=2$$

$$\Rightarrow$$ $$\dfrac{1}{2}=cos x$$

$$\Rightarrow$$ $$x=\cos^{-1}\left(\dfrac{1}{2}\right)$$

$$\Rightarrow$$ $$x=\dfrac{\pi}{3}$$
Substituting value of $$x$$ in ( 1 ) we get,
$$\Rightarrow$$  $$\tan^{-1}\left(2\sin \dfrac{\pi}{3}\right)$$

$$\Rightarrow$$  $$\tan^{-1}\left(2\times \dfrac{\sqrt{3}}{2}\right)$$

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

$$\Rightarrow$$ $$\dfrac{\pi}{3}$$

$$\therefore$$  $$\tan^{-1}\left(2\sin\left(\sec^{-1}(2)\right)\right)=\dfrac{\pi}{3}$$
Question 2
1 / -0

The value of $$\cos ^{ -1 }{ \left\{ \frac { \sqrt { 1-\sin { x } } +\sqrt { 1+\sin { x } } }{ \surd \left( 1-\sin { x } \right) -\surd \left( 1+\sin { x } \right) } \right\} }$$ is $$\left( 0<x<2\pi \right)$$

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

B
$$2\pi -x$$

C
$$-\frac { x}{ 2 } $$

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

Solution

$$ cot^{-1}(\dfrac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}})$$
$$ sin 2x = 2sinx cosx. $$
also, $$ sinx = 2sin \dfrac{x}{2} cos\dfrac{x}{2} ...(1)$$
also, $$ 1+sinx = 2sin\dfrac{x}{2} cos \dfrac{x}{2}+1 $$
since $$ sin^2x+cos^2x = 1 $$
$$ sin^2 \dfrac{x}{2}+cos^2 \dfrac{x}{2} = 1 $$
$$ 1+sinx = sin^2 \dfrac{x}{2}+cos^2 \dfrac{x}{2} + 2sin \dfrac{x}{2} cos \dfrac{x}{2}$$
$$ 1+sinx = (sin\dfrac{x}{2}+cos \dfrac{x}{2})^2$$
$$ \sqrt{1+sinx} = sin\dfrac{x}{2}+cos\dfrac{x}{2} ...(2)$$
Similarly
multiply by -1 in $$ eq^n $$ (1),
$$ -sinx = 2sin\dfrac{x}{2} cos\dfrac{x}{2}$$
Now adding 1 to the above $$ eq^n $$
$$ 1-sinx = 1-2sin \dfrac{x}{2}cos\dfrac{x}{2}$$
$$ \therefore $$ we get $$ \sqrt{1-sinx} = (cos\dfrac{x}{2}-sin\dfrac{x}{2})...(3)$$
$$ cos^{-1}(\dfrac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}})$$
$$ = cos^{-1} (\frac{cos\dfrac{x}{2}-sin\dfrac{x}{2}+sin\dfrac{x}{2}+cos\dfrac{x}{2}}{cos\dfrac{x}{2}-sin\dfrac{x}{2}-sin\dfrac{x}{2}-cos\dfrac{x}{2}})$$
$$ = cos^{-1}(\frac{2cos\dfrac{x}{2}}{-2sin\dfrac{x}{2}}) = cot^{-1}(cot\dfrac{x}{2})$$
$$ = -cot^{-1}(cot\dfrac{x}{2}) (\therefore cotx $$ is an add function)
$$ = -\dfrac{x}{2}$$
Question 3
1 / -0

Annual expenses of A and B are in the ratio $$5:3$$. The saving of Aand B are in the ratio $$1:2$$. Find the expenses of A. given that the income of A is $$Rs. 8000$$ and that of B is $$Rs.9000$$.

A
$$3000$$

B
$$4000$$

C
$$4500$$

D
$$5000$$

Solution

Given

Income of $$A$$ is $$8000$$

Income of $$B$$ is $$9000$$

Expense Ratio is $$5:3$$

Let the common factor be $$x$$

So Expenses of $$A$$ are $$5x$$

Expenses of $$B$$ are $$3x$$

Savings Ratio is $$1:2$$

Let the common factor be $$y$$

So Savings of $$A$$ are $$y$$

Savings of $$B$$ are $$2y$$

According to Question

$$5x+y=8000\cdots(1)$$

$$3x+2y=9000\cdots(2)$$

$$2(1)-(2)$$

$$\implies 10x+2y-3x-2y=16000-9000$$

$$\implies 7x=7000$$

$$\implies x=1000$$

Expenses of $$A$$ is $$5x=5\times 1000=5000$$
Question 4
1 / -0

$$2\cot ^{ -1 }{ 7 } +\cos ^{ -1 }{ \dfrac { 3 }{ 5 } } $$ is equal to

A
$$\cot ^{ -1 }{ \left( \dfrac { 44 }{ 117 } \right) } $$

B
$$cosec ^{ -1 }{ \left( \dfrac { 125 }{ 117 } \right) } $$

C
$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 117 } \right) } $$

D
$$\tan ^{ -1 }{ \left( \dfrac { 44 }{ 125 } \right) } $$

Solution

$$2\cot^{-1}7+\cos^{-1}\left(\dfrac{3}{5}\right)$$

Let $$\cos^{-1}\left(\dfrac{3}{5}\right)=\theta$$

$$\cos\theta =\dfrac{3}{5}$$

$$\cot\theta =\dfrac{3}{4}$$

$$2\cot^{-1}(7)+\cot^{-1}\left(\dfrac{3}{4}\right)$$ $$\left[\because 2\cot^{-1}(x)=\cot^{-1}\left(\dfrac{x^2-1}{2x}\right)\right]$$

$$\cot^{-1}\left(\dfrac{7^2-1}{2\times 7}\right)+\cot^{-1}\left(\dfrac{3}{4}\right)$$

$$\cot^{-1}\left(\dfrac{24}{7}\right)+\cot^{-1}\left(\dfrac{3}{4}\right)$$

$$\cot^{-1}\left(\dfrac{\dfrac{24}{7}\times \dfrac{3}{4}-1}{\dfrac{24}{7}+\dfrac{3}{4}}\right)$$ $$\left[\because \cot^{-1}(a)+\cot^{-1}(b)=\cot^{-1} \left(\dfrac{ab-1}{a+b}\right)\right]$$

$$\cot^{-1}\left(\dfrac{44/28}{117/28}\right)=\cot^{-1}\left(\dfrac{44}{117}\right)$$.
Question 5
1 / -0

$${ cos }^{ -1 }\left( \dfrac { 3+5\cos { x } }{ 5+3\cos { x } } \right) =$$

A
$${ tan }^{ -1 }\left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right)$$

B
$$2\tan ^{ -1 }{ \left(- \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$

C
$$\dfrac { 1 }{ 2 } \tan ^{ -1 }{ \left( 2\tan { \dfrac { x }{ 2 } } \right) }$$

D
$$2\tan ^{ -1 }{ \left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$

Solution

Let $$\cos^{-1}\dfrac{(3+5\cos x)}{(5+3\cos x)}=\alpha$$
$$\dfrac{3+5\cos x}{5+3\cos x}=\cos \alpha$$
$$\because \cos \alpha\dfrac{[1-\tan^2(\dfrac{\alpha}{2})]}{[1+\tan^2(\dfrac{x}{2})]}$$
$$\cos \alpha=5\dfrac{\dfrac{[1-\tan^2(\dfrac{x}{2})]}{[1+\tan^2(\dfrac{x}{2})]}+3}{5+3[\dfrac{1-\tan^2(\dfrac{x}{2})}{1+\tan^2(\dfrac{x}{2})}]}$$
$$\Rightarrow \dfrac{5(1-\tan^2(\dfrac{x}{2}))+3(1+\tan^2(\dfrac{x}{2}))}{5(1+\tan^2(\dfrac{x}{2}))+3(1-\tan^2(\dfrac{x}{2}))}$$
$$\Rightarrow \dfrac{5-5\tan^2(\dfrac{x}{2})+3+3\tan^2(\dfrac{x}{2})}{5+5\tan^2(\dfrac{x}{2})+3-3\tan^2(\dfrac{x}{2})}$$
$$\Rightarrow \dfrac{8-2\tan^2(\dfrac{x}{2})}{8+2\tan^2(\dfrac{x}{2})}$$
$$\dfrac{4-\tan^2(\dfrac{x}{2})}{4+\tan^2(\dfrac{x}{2})}$$

$$\therefore \sin \alpha=\sqrt{1-\cos^2\alpha}$$
$$\Rightarrow \sin \alpha=\sqrt{1-\left(\dfrac{4-\tan^2\left(\dfrac{x}{2}\right)}{4+\tan^2\left(\dfrac{x}{2}\right)}\right)^2}$$

$$\Rightarrow \sqrt{\dfrac{\left[4+\tan^2(\dfrac{x}{2}\right)^2]\left[4+\tan^2(\dfrac{x}{2})\right]}{4+\tan^2\left(\dfrac{x}{2}\right)}^2}^2$$

$$=\sqrt{\dfrac{4+\tan^2\left(\dfrac{x}{2}\right)-4+\tan^2\left(\dfrac{x}{2}\right)\left(4+\tan^2(\dfrac{x}{2}+4-\tan^2\right)\left(\dfrac{x}{2})\right)}{\left[4+\tan^2(\dfrac{x}{2})\right]^2}}$$

$$\Rightarrow \dfrac{\sqrt{2\tan^2(\dfrac{x}{2}).8}}{4+\tan^2(\dfrac{x}{2})}\Rightarrow \dfrac{4\tan(\dfrac{x}{2})}{4\tan^2(\dfrac{n}{2})}$$

$$\because \tan\left(\dfrac{x}{2}\right)=\dfrac{1-\cos \alpha}{\sin \alpha}$$
$$\therefore \dfrac{1-\left(\dfrac{4-\tan^2(\dfrac{n}{2})}{4+\tan^2(\dfrac{n}{2})}\right)}{\left(\dfrac{4\tan(\dfrac{n}{2})}{4+\tan^2(\dfrac{n}{2})}\right)}\Rightarrow \dfrac{4+\tan^2\left(\dfrac{n}{2}-4+\tan^2\right)\left(\dfrac{n}{2}\right)}{4\tan \left(\dfrac{n}{2}\right)}$$

$$\tan\left(\dfrac{x}{2}\right)\Rightarrow \dfrac{2\tan^2\left(\dfrac{x}{2}\right)}{2^4\tan\left(\dfrac{x}{2}\right)}\Rightarrow \dfrac{1}{2}\tan\left(\dfrac{x}{2}\right)$$

$$\dfrac{\alpha}{2}=\tan^{-1}\left[\dfrac{1}{2}\tan(\dfrac{x}{2})\right]$$

$$\boxed{\cos^{-1}\left(\dfrac{5 \cos x+3}{5+3\cos x}\right)\Rightarrow 2\tan^{-1}\left[\dfrac{1}{2}\tan(\dfrac{n}{2})\right]}$$
So, option (b) is correct.
Question 6
1 / -0

The value of $$\cos\left\{\tan^{-1}\left(\tan \dfrac{15\pi}{4}\right)\right\}$$ is?

A
$$\dfrac{1}{\sqrt{2}}$$

B
$$-\dfrac{1}{\sqrt{2}}$$

C
$$1$$

D
None of these

Solution

Given

$$\cos \left(\tan ^{-1} \left(\tan \dfrac {15\pi}4 \right)\right)$$

$$\implies \cos \left(\tan ^{-1} \left(\tan \left(4\pi -\dfrac {\pi}4 \right)\right)\right)$$

$$\implies \cos \left(\tan ^{-1} \left(\tan \left(-\dfrac {\pi}4 \right)\right)\right)$$

$$\bigg(\tan (-x)=-\tan x\bigg)$$

$$\implies \cos \left(-\dfrac {\pi}4 \right)$$

$$\implies \dfrac 1{\sqrt 2}$$
Question 7
1 / -0

If the number $$93215x2$$ is completely divisible by $$11$$, then $$x$$ is equal to

A
$$2$$

B
$$3$$

C
$$1$$

D
$$4$$

Solution

Given number is $$93215x2$$

The number can be represented as $$9321502+10x$$

Dividing $$9321502$$ leaves $$3$$ as remainder

So the expression $$10x+3$$ should be divisible by $$11$$ to make the whole number to be divisible by $$11$$

$$\implies 10x+3=11m(say)$$

$$\implies x=\dfrac{11m-3}{10} $$

The expression gives an integer value at $$m=3$$

$$\implies x=\dfrac{33-3}{10}$$

$$\implies x=3$$
Question 8
1 / -0

If $$\sin^{-1}\left(x-\dfrac {x^{2}}{2}+\dfrac {x^{3}}{4}-...\infty \right)+\cos^{-1}\left(x^{2}-\dfrac {x^{4}}{2}+\dfrac {x^{6}}{4}-...\infty \right)=\dfrac {\pi}{2}$$ for $$0 < |x| < \sqrt {2}$$,then $$x$$ equal

A
$$\dfrac {1}{2}$$

B
$$1$$

C
$$\dfrac {-1}{2}$$

D
$$-1$$

Solution

As we know that
Sum of an infinite $$GP$$ is $$\dfrac{a}{1-r}$$ for $$|r|<1$$

So $$x-\dfrac{x^2}{2}+\dfrac{x^3}{4}-\cdots\infty=\dfrac{x}{1+\dfrac{x}{2}}$$

And also $$x^2-\dfrac{x^2}{2}+\dfrac{x^6}{4}-\cdots\infty=\dfrac{x^2}{1+\dfrac{x^2}{2}}$$

So $$\text{sin}^{-1}\bigg(x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots\infty\bigg)+\text{cos}^{-1}\bigg(x^2-\frac{x^4}{4}+\frac{x^6}{4}-\cdots\infty\bigg)=\dfrac{\pi}{2}$$

$$\implies \text{sin}^{-1}\bigg(\dfrac{x}{1+\frac{x}{2}}\bigg)=\dfrac{\pi}{2}-\text{cos}^{-1}\bigg(\dfrac{x^2}{1+\frac{x^2}{2}}\bigg)$$

$$\implies \text{sin}^{-1}\bigg(\dfrac{2 x}{x+2}\bigg)=\text{sin}^{-1}\bigg(\dfrac{2 x^2}{x^2+2}\bigg)$$

$$\implies \dfrac{1}{x+2}=\dfrac{x}{x^2+2}$$

$$\implies x^2+2 x=x^2+2$$

$$\implies 2 x=2$$

$$\implies x=1$$
Question 9
1 / -0

The value of $$\cot^{-1}{3}+sec^{-1}{\dfrac{\sqrt{5}}{2}}$$ is

A
$$\dfrac{\pi}{4}$$

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{\pi}{2}$$

D
None of these

Solution

$$\cot^{-1}(3)+\sec^{-1}\sqrt{5}/2$$
$$\tan^{-1}\left(\dfrac{1}{3}\right)+\tan^{-1}\left(\dfrac{1}{2}\right)$$
$$\tan^{-1}\left(\dfrac{\dfrac{5}{6}}{1-\dfrac{1}{6}}\right)=\tan^{-1}(1)$$
$$=\dfrac{\pi}{4}$$
Question 10
1 / -0

The value of $$\tan^{-1}\dfrac {1}{2}+\tan^{-1}\dfrac {1}{3}$$ is

A
$$\dfrac {\pi}{4}$$

B
$$\dfrac {\pi}{6}$$

C
$$\dfrac {\pi}{3}$$

D
$$0$$

Solution

$$\tan^{-1}\left(\dfrac{1}{2}\right)+\tan^{-1}\left(\dfrac{1}{3}\right)$$

use formula
$$\tan^{-1}{a}+\tan^{-1}{b}$$ = $$\tan^{-1} (\dfrac{a+b}{1-ab})$$

$$\Rightarrow \tan^{-1}\left(\dfrac{5}{6-1}\right)=\tan^{-1}\left(\dfrac{5}{5}\right)$$

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

$$=\dfrac{\pi}{4}$$.

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Inverse Trigonometric Functions Test - 36

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Inverse Trigonometric Functions Test - 36

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SHARING IS CARING

Question 1

$$\dfrac {\pi}{6}$$

$$\dfrac {\pi}{4}$$

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

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

Solution

Question 2

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

$$2\pi -x$$

$$-\frac { x}{ 2 } $$

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

Solution

Question 3

$$3000$$

$$4000$$

$$4500$$

$$5000$$

Solution

Question 4

$$\cot ^{ -1 }{ \left( \dfrac { 44 }{ 117 } \right) } $$

$$cosec ^{ -1 }{ \left( \dfrac { 125 }{ 117 } \right) } $$

$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 117 } \right) } $$

$$\tan ^{ -1 }{ \left( \dfrac { 44 }{ 125 } \right) } $$

Solution

Question 5

$${ tan }^{ -1 }\left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right)$$

$$2\tan ^{ -1 }{ \left(- \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$

$$\dfrac { 1 }{ 2 } \tan ^{ -1 }{ \left( 2\tan { \dfrac { x }{ 2 } } \right) }$$

$$2\tan ^{ -1 }{ \left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$

Solution

Question 6

$$\dfrac{1}{\sqrt{2}}$$

$$-\dfrac{1}{\sqrt{2}}$$

$$1$$

None of these

Solution

Question 7

$$2$$

$$3$$

$$1$$

$$4$$

Solution

Question 8

$$\dfrac {1}{2}$$

$$1$$

$$\dfrac {-1}{2}$$

$$-1$$

Solution

Question 9

$$\dfrac{\pi}{4}$$

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

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

None of these

Solution

Question 10

$$\dfrac {\pi}{4}$$

$$\dfrac {\pi}{6}$$

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

$$0$$

Solution

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