Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

$$\tan (2\cos ^{-1}\frac 35)=$$_____

A
$$\frac 83$$

B
$$\frac {24}{25}$$

C
$$\frac 7{25}$$

D
$$\frac {-24}7$$

Solution

Given,

$$\tan \left(2\cos ^{-1}\left(\frac{3}{5}\right)\right)$$

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

$$=\dfrac{2\tan \left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}{1-\tan ^2\left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}$$

as $$\tan \left(\cos ^{-1}\left(x\right)\right)=\frac{\sqrt{1-x^2}}{x}$$

$$=\dfrac{2\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)}{1-\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)^2}$$

$$=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\left(\frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}\right)^2}$$

$$=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\frac{4^2}{3^2}}$$

$$=\dfrac{\frac{8}{3}}{1-\frac{4^2}{3^2}}$$

$$=\dfrac{8}{3\left(1-\frac{16}{9}\right)}$$

$$=-\dfrac{8}{\frac{7}{3}}$$

$$=-\dfrac{24}{7}$$
Question 2
1 / -0

The solution set of the equation$$2 cos^{ -1 } x = cot^{ -1 } \left(\dfrac { 2x^{ 2 } - 1 }{ 2x \sqrt { 1 x^{ 2 } } }\right) $$ is

A
$$\left(0, 1\right)$$

B
$$\left(-1, 1\right)-{ 0 }$$

C
$$\left(-1, 0\right)$$

D
$$\left[-1, 1\right]$$

Solution
Question 3
1 / -0

If $$\;\sin {\;^{ - 1}}\dfrac{1}{x} = 2\;{\tan ^{ - 1}}\dfrac{1}{7} + {\cos ^{ - 1}}\dfrac{3}{5}$$, then $$x =$$ ___

A
$$\dfrac {24}{117}$$

B
$$\dfrac {7}{3}$$

C
$$\dfrac {125}{117}$$

D
None of these

Solution

$$2\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$
$$=\tan ^{-1}\left(\dfrac{1}{7}\right)+\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{1}{7}+\dfrac{1}{7}}{1-\dfrac{1}{7}\times \dfrac{1}{7}}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$ $$\left[\because \tan ^{-1}+\tan ^{-1}y \tan ^{-1x}=\tan ^{-1}\left(\dfrac{x+y}{1-xy}\right)\right]$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{2}{7}}{1-\dfrac{1}{49}}\right)+\theta$$
$$=\tan ^{-1}\left(\dfrac{7}{24}\right)+\tan ^{-1}\left(\dfrac{4}{3}\right)$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{7}{24}+\dfrac{4}{3}}{1-\dfrac{7}{24}\times \dfrac{4}{3}}\right)$$
$$=\tan ^{-1}\left(\dfrac{117}{44}\right)$$
$$\because \sin ^{-1}\left(\dfrac{1}{x}\right)=\tan ^{-1}\left(\dfrac{117}{44}\right)$$
Let $$\theta =\tan ^{-1}\left(\dfrac{117}{44}\right)$$
$$\therefore \tan \theta=\dfrac{117}{44}$$
$$\therefore \sin \theta=\dfrac{117}{\sqrt{117^2+44^2}}=\dfrac{117}{125}$$
$$\therefore \theta =\sin ^{-1}\left(\dfrac{117}{125}\right)$$
$$\therefore \sin ^{-1}\left(\dfrac{1}{x}\right)=\sin ^{-1}\left(\dfrac{117}{125}\right)$$
On comparing
$$\dfrac{1}{x}=\dfrac{117}{125}$$
$$\boxed{\therefore x=\dfrac{125}{117}}$$
Question 4
1 / -0

If $$\alpha =\cos^{-1}\left(\dfrac{3}{5}\right),\beta=\tan^{-1}\left(\dfrac{1}{3}\right)$$ where $$0<\alpha,beta <\dfrac{\pi}{2},$$then $$\alpha -\beta $$ is equal to :

A
$$\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

B
$$\tan^{-1}\left(\dfrac{9}{14}\right)$$

C
$$\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

D
$$\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

Solution

$$\cos \alpha=\dfrac{3}{5},\tan \beta =\dfrac{1}{5}$$
$$\Rightarrow \tan \alpha =\dfrac{4}{3}$$
$$\Rightarrow (\alpha-\beta)=\dfrac{\dfrac{4}{3}-\dfrac{1}{3}}{1+\dfrac{4}{3}.\dfrac{1}{3}}=\dfrac{9}{3}$$
$$\Rightarrow \sin(\alpha-\beta)=\dfrac{9}{5\sqrt{10}}$$
$$\Rightarrow \alpha-\beta =\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$
Question 5
1 / -0

If $$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha $$ where $$-1-1\le x\le 1,-2\le y\le 2,x\le \cfrac { y }{ 2 } $$ then for all $$4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }$$ is equal to

A
$$4\sin ^{ 2 }{ \alpha } -2{ x }^{ 2 }{ y }^{ 2 }$$

B
$$4\cos ^{ 2 }{ \alpha } +2{ x }^{ 2 }{ y }^{ 2 }$$

C
$$4\sin ^{ 2 }{ \alpha } $$

D
$$2\sin ^{ 2 }{ \alpha } $$

Solution

$$\cos { \left( \cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } \right) } =\cos { \alpha } $$
$$\cos { \alpha } \Rightarrow x\times \cfrac { y }{ 2 } +\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { y }^{ 2 } }{ 4 } } \Rightarrow { \left( \cos { \alpha } -\cfrac { xy }{ 2 } \right) }^{ 2 }=\left( 1-{ x }^{ 2 } \right) \left( 1-\cfrac { { y }^{ 2 } }{ 4 } \right) $$
$${ x }^{ 2 }+\cfrac { { y }^{ 2 } }{ 4 } -xy\cos { \alpha } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha } \quad $$
Question 6
1 / -0

The value of $$tan \left (\cos^{-1} \dfrac {3}{5} + \tan^{-1} \dfrac {1}{4}\right )$$ is

A
$$\dfrac {19}{8}$$

B
$$\dfrac {8}{19}$$

C
$$\dfrac {19}{12}$$

D
$$\dfrac {3}{4}$$

Solution

Given, $$\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$

Let, $${\cos}^{-1}{\dfrac{3}{5}}=a$$

$$\Rightarrow\,\cos{a}=\dfrac{3}{5}$$

$$\Rightarrow\,{\sin}^{2}{a}=1-{\cos}^{2}{a}=1-\dfrac{9}{25}=\dfrac{25-9}{25}=\dfrac{16}{25}$$

$$\therefore\,\sin{a}=\dfrac{4}{5}$$

$$\Rightarrow\,\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}$$

$$\Rightarrow\,a={\tan}^{-1}{\dfrac{4}{3}}$$

$$\therefore\,\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$

$$=\tan{\left(a+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$ where $$a={\cos}^{-1}{\dfrac{3}{5}}$$

$$=\tan{\left({\tan}^{-1}{\dfrac{4}{3}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$ where $$a={\cos}^{-1}{\dfrac{3}{5}}={\tan}^{-1}{\dfrac{4}{3}}$$

We know that $${\tan}^{-1}{A}+{\tan}^{-1}{B}={\tan}^{-1}{\left(\dfrac{A+B}{1-AB}\right)}$$

$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{4}{3}+\dfrac{1}{4}}{1-\dfrac{4}{3}\times\dfrac{1}{4}}\right)}\right)}$$

$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{16+3}{12}}{1-\dfrac{4}{12}}\right)}\right)}$$

$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{19}{12}}{\dfrac{12-4}{12}}\right)}\right)}$$

$$=\tan{\left({\tan}^{-1}{\dfrac{19}{8}}\right)}$$

$$=\dfrac{19}{8}$$
Question 7
1 / -0

The value of $$\tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{8}$$ is ___________.

A
$$\dfrac{11\pi}{5}$$

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

C
$$\pi$$

D
$$\dfrac{3\pi}{4}$$

Solution

$$\tan^{-1} \left (\dfrac{1}{3} \right ) + \tan^{-1} \left (\dfrac{1}{5} \right ) + \tan^{-1} \left (\dfrac{1}{7} \right ) + \tan^{-1} \left (\dfrac{1}{8} \right ) $$

$$ = \tan^{-1} \left (\dfrac{\dfrac{1}{3} + \dfrac{1}{5}}{1 - \dfrac{1}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{1}{7} + \dfrac{1}{8}}{1 - \dfrac{1}{56}} \right ) $$

$$= \tan^{-1} \left (\dfrac{\dfrac{8}{15}}{\dfrac{14}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{15}{56}}{\dfrac{55}{56}} \right ) $$

$$ = \tan^{-1} \left (\dfrac{8}{14} \right ) + \tan^{-1} \left (\dfrac{15}{55} \right )$$

$$ = \tan^{-1} \left (\dfrac{4}{7} \right ) + \tan^{-1} \left (\dfrac{3}{11} \right )$$

$$= \tan^{-1} \left (\dfrac{\dfrac{4}{7} + \dfrac{3}{11}}{1 - \dfrac{12}{77}} \right ) $$

$$= \tan^{-1} \left (\dfrac{\dfrac{65}{77}}{\dfrac{65}{77}} \right )$$

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

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

$$= \dfrac{\pi}{4}$$
Question 8
1 / -0

$$\tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=$$ ?

A
$$\dfrac {7}{17}$$

B
$$\dfrac {-7}{17}$$

C
$$\dfrac {7}{12}$$

D
$$\dfrac {-7}{12}$$

Solution

Let $$2\tan^{-1}\dfrac {1}{5}=\theta$$.

Then, $$\tan^{-1}\dfrac {1}{5}=\dfrac {1}{2}\theta \Rightarrow \tan \dfrac {\theta}{2} =\dfrac {1}{5}$$

$$\therefore \ \tan \theta =\dfrac {2\tan \dfrac {1}{2}\theta}{1-\tan^2 \dfrac {1}{2}\theta}=\dfrac {\left(2\times \dfrac {1}{5}\right)}{\left(1-\dfrac {1}{25}\right)}=\left (\dfrac {2}{5}\times \dfrac {25}{24}\right)=\dfrac {5}{12}$$

$$\therefore \ \tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=\tan \left (\theta -\dfrac {\pi}{4}\right)=\dfrac {\tan \theta -\tan \dfrac {\pi}{4}}{1+\tan \theta \cdot \tan \dfrac {\pi}{4}}=\dfrac {\left (\dfrac {5}{12}-1\right)}{\left(1+\dfrac {5}{12}\times 1\right)}=\dfrac {\left (\dfrac {-7}{12}\right)}{\left (\dfrac {17}{12}\right)}=\dfrac {-7}{17}$$
Question 9
1 / -0

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

A
$$\dfrac {3}{4}$$

B
$$\dfrac {4}{5}$$

C
$$\dfrac {3}{5}$$

D
$$none\ of\ these$$

Solution
Question 10
1 / -0

The value of $$\sin \left (\sin^{-1}\dfrac {1}{2}+\cos^{-1}\dfrac {1}{2}\right)=$$?

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$none\ of\ these$$

Solution

As we know that, $$\sin^{-1} x +\cos^{-1}x=\dfrac {\pi}{2},\,x\in[-1,1]$$

$$ \sin \left (\sin^{-1}\dfrac {1}{2}+\cos^{-1}\dfrac {1}{2}\right)=\sin \dfrac {\pi}{2}=1$$

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 44

Result

Inverse Trigonometric Functions Test - 44

Score

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Rank

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SHARING IS CARING

Question 1

$$\frac 83$$

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

$$\frac 7{25}$$

$$\frac {-24}7$$

Solution

Question 2

$$\left(0, 1\right)$$

$$\left(-1, 1\right)-{ 0 }$$

$$\left(-1, 0\right)$$

$$\left[-1, 1\right]$$

Solution

Question 3

$$\dfrac {24}{117}$$

$$\dfrac {7}{3}$$

$$\dfrac {125}{117}$$

None of these

Solution

Question 4

$$\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

$$\tan^{-1}\left(\dfrac{9}{14}\right)$$

$$\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

$$\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$

Solution

Question 5

$$4\sin ^{ 2 }{ \alpha } -2{ x }^{ 2 }{ y }^{ 2 }$$

$$4\cos ^{ 2 }{ \alpha } +2{ x }^{ 2 }{ y }^{ 2 }$$

$$4\sin ^{ 2 }{ \alpha } $$

$$2\sin ^{ 2 }{ \alpha } $$

Solution

Question 6

$$\dfrac {19}{8}$$

$$\dfrac {8}{19}$$

$$\dfrac {19}{12}$$

$$\dfrac {3}{4}$$

Solution

Question 7

$$\dfrac{11\pi}{5}$$

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

$$\pi$$

$$\dfrac{3\pi}{4}$$

Solution

Question 8

$$\dfrac {7}{17}$$

$$\dfrac {-7}{17}$$

$$\dfrac {7}{12}$$

$$\dfrac {-7}{12}$$

Solution

Question 9

$$\dfrac {3}{4}$$

$$\dfrac {4}{5}$$

$$\dfrac {3}{5}$$

$$none\ of\ these$$

Solution

Question 10

$$0$$

$$1$$

$$-1$$

$$none\ of\ these$$

Solution

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