Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

$$\frac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}=$$

A
4

B
3

C
2

D
1

Solution

$$\textbf{Step-1: Rewrite the Numerator observing Denominator}$$
$$\textbf{& apply multiple - angles of trigonometric function.}$$

$$\text{We have,}$$

$$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$$ $$\text{.....(i)}$$

$$\text{Now,}\\$$
$$\cos^{-1} (1- 2 (\dfrac{2}{7})^2)$$

$$\text{Put}$$ $$\sin \theta = \dfrac{2}{7}$$

$$\therefore$$ $$\cos^{-1}(1- 2 \sin^2 \theta)$$

$$\cos^{-1} (\cos 2\theta)$$ $$\boldsymbol{[\cos 2 \theta = 1 - 2 \sin^{2} \theta]}$$

$$ = 2\theta$$

$$\textbf{Step-2: Use the above results & get the required unknown.}$$

$$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$$ $$= \dfrac{2 \theta}{\theta} = 2$$ $$[\textbf{using (i)}]$$

$$\textbf{Hence, option - C is the answer}$$
Question 2
1 / -0

The value of $$tan(\frac { 1 }{ 2 } { cos }^{ -1 }(\frac { \sqrt { 5 } }{ 3 } ))$$ is

A
$$\frac { 3+\sqrt { 5 } }{ 2 } $$

B
$$3-\sqrt { 5 } $$

C
$$\frac { 1 }{ 2 } (3-\sqrt { 5 } )$$

D
none of these

Solution

Given,

$$y=\tan \left [ \dfrac{1}{2}\cos ^{-1}\left ( \dfrac{\sqrt{5}}{3} \right ) \right ]$$

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

$$\Rightarrow \cos x=\dfrac{\sqrt{5}}{3} $$

$$\therefore y=\tan \dfrac{1}{2}x$$

$$y=\tan \dfrac{x}{2}$$

$$y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$$

$$=\sqrt{\dfrac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}$$

$$=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}$$

rationalizing the factor, we get,

$$y=\dfrac{3-\sqrt{5}}{2}$$
Question 3
1 / -0

If tan (x + y) = 33 and x = $${ tan }^{ -1 }3$$, then y will be

A
0.3

B
$${ tan }^{ -1 }(1.3)$$

C
$${ tan }^{ -1 }(0.3)$$

D
$${ tan }^{ -1 }(\frac { 1 }{ 18 } )$$

Solution

Given,

$$\tan (x+y)=33$$

$$x=\tan ^{-1}3$$

$$\Rightarrow \tan x=3$$

Formula,

$$\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$$

$$33=\dfrac{3+\tan y}{1-(3)\tan y}$$

$$33(1-3\tan y)=3+\tan y$$

$$33-99\tan y=3+\tan y$$

$$30=100\tan y$$

$$\tan y=\dfrac{30}{100}=0.3$$

$$\therefore y=\tan ^{-1}0.3$$
Question 4
1 / -0

$$\tan { ^{ -1 }\left( 3/5 \right) } +\tan { ^{ -1 }\left( 1/4 \right) } =$$

A
$$0$$

B
$$\pi /4$$

C
$$3n/4$$

D
None of these

Solution

given, $${\tan}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}$$

$$={\tan}^{-1}{\left(\dfrac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\times\frac{1}{4}}\right)}$$

$$={\tan}^{-1}{\left(\dfrac{\frac{12+5}{20}}{1-\frac{3}{20}}\right)}$$

$$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{20-3}{20}}\right)}$$

$$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{17}{20}}\right)}$$

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

$$=\dfrac{\pi}{4}$$
Question 5
1 / -0

If $$sin^{-1}(\dfrac{1}{3}) + sin^{-1}(\dfrac{2}{3}) = sin^{-1} x$$, then x is equal to

A
$$0$$

B
$$\dfrac{\sqrt{5}+4\sqrt{2}}{9}$$

C
$$\dfrac{5\sqrt{2}-4\sqrt{5}}{9}$$

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

Solution

$$\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{2}{3}\\\sin ^{-1}\left\{\dfrac{1}{3}\sqrt{1-\left(\dfrac{2}{3}\right)^2}+\dfrac{2}{3}\sqrt{1-\left(\dfrac{1}{3}\right)^2}\right\}\\\sin^{-1}\left\{\dfrac{1}{3}\left(\dfrac{\sqrt5}{3}\right)+\dfrac{2}{3}\left(\dfrac{\sqrt8}{3}\right)\right\}\\\sin^{-1}\left\{\dfrac{\sqrt5+4\sqrt2}{9}\right\}\\x=\dfrac{\sqrt5+4\sqrt2}{9}$$
Question 6
1 / -0

The value of $$\sin ^{-1}(\sin 5\frac {\pi}{3})=$$

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

B
$$\frac {\pi}3$$

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

D
$$\frac {3\pi}3$$

Solution

Given,

$$\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )$$

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

$$\sin \theta =\left (\sin 5\dfrac{\pi }{3} \right )=\sin \left ( 2\pi - \dfrac{\pi }{3}\right )$$

$$\Rightarrow \sin \theta =-\sin \dfrac{\pi }{3}$$

$$\therefore \theta =-\dfrac{\pi }{3}$$
Question 7
1 / -0

The value of $$\displaystyle sec\left [ sin^{-1}\left (sin \dfrac{50\pi }{9} \right ) + cos^{-1}cos\left ( \dfrac{31\pi }{9} \right ) \right ]$$ is equal to

A
$$sec\dfrac{10\pi}{9}$$

B
$$sec \ 9{\pi}$$

C
$$-1$$

D
$$1$$

Solution

Given,

$$\sec \left [ \sin ^{-1}\left ( \sin \dfrac {50\pi }{9} \right ) +\cos ^{-1}\left ( \cos \dfrac {31\pi }{9} \right ) \right ]$$

$$=\sec \left [ \dfrac {50\pi }{9} + \dfrac {31\pi }{9}\right ]$$

$$=\sec \dfrac {81\pi }{9}$$

$$=\sec 9\pi$$
Question 8
1 / -0

$${ \cos }^{ -1 }\left[ \cos \left( 2{ \cot }^{ -1 }\left( \sqrt { 2 } -1 \right) \right) \right]$$ is equal to

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

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

C
$$\pi/4$$

D
$$3\pi/4$$

Solution
Question 9
1 / -0

The value of $$\sin^{-1}(\cos (\log_{2}(4\alpha -44)))$$ is

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

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

C
$$0$$

D
$$1$$
Question 10
1 / -0

The number of solutions of the equation $$3\cos^{-1}x-\pi x-\dfrac {\pi}{2}=0$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$infinite$$

Solution

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Inverse Trigonometric Functions Test - 43

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Inverse Trigonometric Functions Test - 43

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SHARING IS CARING

Question 1

4

3

2

1

Solution

Question 2

$$\frac { 3+\sqrt { 5 } }{ 2 } $$

$$3-\sqrt { 5 } $$

$$\frac { 1 }{ 2 } (3-\sqrt { 5 } )$$

none of these

Solution

Question 3

0.3

$${ tan }^{ -1 }(1.3)$$

$${ tan }^{ -1 }(0.3)$$

$${ tan }^{ -1 }(\frac { 1 }{ 18 } )$$

Solution

Question 4

$$0$$

$$\pi /4$$

$$3n/4$$

None of these

Solution

Question 5

$$0$$

$$\dfrac{\sqrt{5}+4\sqrt{2}}{9}$$

$$\dfrac{5\sqrt{2}-4\sqrt{5}}{9}$$

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

Solution

Question 6

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

$$\frac {\pi}3$$

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

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

Solution

Question 7

$$sec\dfrac{10\pi}{9}$$

$$sec \ 9{\pi}$$

$$-1$$

$$1$$

Solution

Question 8

$$\sqrt{2}-1$$

$$1-\sqrt{2}$$

$$\pi/4$$

$$3\pi/4$$

Solution

Question 9

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

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

$$0$$

$$1$$

Question 10

$$0$$

$$1$$

$$2$$

$$infinite$$

Solution

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