Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 46

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Question 1
1 / -0

The value of
$$\displaystyle sin^{-1} \left \{ (sin\,\pi/3) \dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \} - cos^{-1} \left \{ cos\pi /6 \,\dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \} $$
$$ \left (where \,\dfrac{k}{2} < x < 2k , \,k > 0 \right ) $$ is

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

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

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

D
none of these

Solution
Question 2
1 / -0

$$cos^{-1}(cos(\dfrac{5\pi}{4}))$$ is given by

A
$$\dfrac{5\pi}{4}$$

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

C
$$\dfrac{-\pi}{4}$$

D
none of these

Solution

$$cos^{-1}\bigg(cos\dfrac{5\pi}{4}\bigg)=cos^{-1}\Bigg(cos\bigg(2\pi-\dfrac{5\pi}{4}\bigg)\Bigg)=cos^{-1}\Bigg(cos\bigg(\dfrac{3\pi}{4}\bigg)\Bigg)=\dfrac{3\pi}{4}$$
Question 3
1 / -0

The value $$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$$ is equal to

A
$$\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$$

B
$$\cos^{-1}\Bigg[\dfrac{b\cos\theta+a}{a\cos\theta+b}\Bigg]$$

C
$$\cos^{-1}\Bigg[\dfrac{a\cos\theta}{b\cos\theta+a}\Bigg]$$

D
$$\cos^{-1}\Bigg[\dfrac{b\cos\theta}{a\cos\theta+b}\Bigg]$$

Solution

$$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$$=$$cos^{-1}\Bigg[\dfrac{1-\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}{1+\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}\Bigg]$$

$$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$$=$$cos^{-1}\Bigg[\dfrac{(a+b)-(a-b)\tan^2\dfrac{\theta}{2}}{(a+b)+{a-b}\tan^2\dfrac{\theta}{2}}\Bigg]$$

$$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$$=$$cos^{-1}\Bigg[\dfrac{a(1-\tan^2\dfrac{\theta}{2})+b(1+\tan^2\dfrac{\theta}{2})}{a(1+\tan^2\dfrac{\theta}{2})+b(1-\tan^2\dfrac{\theta}{2})}\Bigg]$$

$$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$$=$$cos^{-1}\Bigg[\dfrac{\dfrac{a(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}+b}{a+b\dfrac{(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}}\Bigg]$$

$$=\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$$
Question 4
1 / -0

If f(x)=$$sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg),-\dfrac{1}{2}\leq x\leq1$$, then f(x) is equal to

A
$$\sin^{-1}\bigg(\dfrac{1}{2}\bigg)-sin^{-1}x$$

B
$$\sin^{-1}x-\dfrac{\pi}{6}$$

C
$$\sin^{-1}x+\dfrac{\pi}{6}$$

D
none of these

Solution

Let $$x=\sin\theta$$ where $$\dfrac{-1}{2}\leq x\leq1\implies\,\dfrac{-\pi}{6}\leq\theta\leq\dfrac{\pi}{2}$$

Then $$f(x)=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg)$$

$$=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}\sin\theta-\dfrac{1}{2}\cos\theta\Bigg)$$

$$=sin^{-1}\Bigg(\sin\bigg(\theta-\dfrac{\pi}{6}\bigg)\Bigg)$$

$$=\theta-\dfrac{\pi}{6}=sin^{-1}x-\dfrac{\pi}{6}$$
Question 5
1 / -0

If $$sin^{-1}\dfrac{5}{x}+sin^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}$$, then x is equal to

A
$$\dfrac{7}{13}$$

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

C
$$13$$

D
$$\dfrac{13}{7}$$

Solution

Put $$sin^{-1}\dfrac{5}{x}=A\implies \dfrac{5}{x}=\sin A$$

$$\sin^{-1}\dfrac{12}{x}=B\implies \dfrac{12}{x}=\sin B \implies A+B=\dfrac{\pi}{2}$$

$$\implies\sin A=\sin(\dfrac{\pi}{2}-B)=\cos B=\sqrt{1-sin^{2}B}$$

$$\implies\dfrac{5}{x}=\sqrt{1-\dfrac{144}{x^2}}\implies \dfrac{169}{x^2}=1$$

$$\implies x^2=169 \implies x=13$$ [$$x = -13$$ does not satisfy the given equation]
Question 6
1 / -0

If $$3\sin^{-1}\bigg(\dfrac{2x}{1+x^2}\bigg)-4\cos^{-1}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)+2tan^{-1}\bigg(\dfrac{2x}{1-x^2}\bigg)=\dfrac{\pi}{3}$$, where $$|x|<1$$. then x is equal to

A
$$\dfrac{1}{\sqrt{3}}$$

B
$$-\dfrac{1}{\sqrt{3}}$$

C
$${\sqrt{3}}$$

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

E
$$\dfrac{\sqrt{3}}{2}$$

Solution

Put $$x=\tan\theta$$
$$\therefore\,3\sin^{-1}\bigg(\dfrac{2\tan\theta}{1+\tan^2\theta}\bigg)-4\cos^{-1}\bigg(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\bigg)+2\tan^{-1}\bigg(\dfrac{2\,\tan\theta}{1-\tan^2\theta}\bigg)=\dfrac{\pi}{3}$$

$$\implies\,3\sin^{-1}(\sin2\theta)-4\cos^{-1}(\cos2\theta)+2\tan^{-1}(\tan2\theta)=\dfrac{\pi}{3}$$
$$\implies\,3(2\theta)-4(2\theta)+2(2\theta)=\dfrac{\pi}{3}\,\implies2\theta=\dfrac{\pi}{3}\implies\theta=\dfrac{\pi}{6}\implies\,x=\tan \dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$$
Question 7
1 / -0

If $$x\in\bigg(\dfrac{-\pi}{2},\dfrac{\pi}{2}\bigg)$$, then the value $$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$$ is

A
$$\dfrac{x}{2}$$

B
$$2x$$

C
$$3x$$

D
$$x$$

Solution

$$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$$=$$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)$$+$$\tan^{-1}\Bigg(\dfrac{\dfrac{6\tan x}{1+tan^{2}x}}{5+3\dfrac{1-tan^{2}x}{1+tan^{2}x}}\Bigg)$$

$$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{6\tan x}{8+2\tan^2x}\Bigg)$$

$$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\tan x}{4+\tan^2x}\Bigg)$$

$$=\tan^{-1}\Bigg(\dfrac{\dfrac{\tan x}{4}+\dfrac{3\tan x}{4+\tan^2x}}{1-\dfrac{3\tan^2x}{4(4+\tan^2x)}}\Bigg)$$

$$=\tan^{-1}\Bigg(\dfrac{16\tan x+tan^{3}x}{16+tan^2x}\Bigg)$$

$$=\tan^{-1}(\tan x)=x$$
Question 8
1 / -0

The value of $$sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$$ is

A
$$0$$

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

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

D
none of these

Solution

We have $$sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$$

= $$\sin^{-1}\Bigg(\cot\bigg(sin^{-1}\bigg(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\bigg)+cos^{-1}\dfrac{\sqrt{3}}{2}+\cos^{-1}\dfrac{1}{\sqrt{2}}\bigg)\Bigg)$$

= $$\sin^{-1}[\cot(15^{\circ}+30^{\circ}+45^{\circ})]$$

= $$\sin^{-1}[\cot(90^{\circ})]$$

= $$\sin^{-1}(0)=0$$
Question 9
1 / -0

If $$\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x$$, then x is equal to

A
$$1$$

B
$$\sqrt{3}$$

C
$$\dfrac{1}{\sqrt{3}}$$

D
none of these

Solution

We have $$\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x$$

$$\implies\,\tan^{-1}\Bigg[\dfrac{1-\tan\theta}{1+\tan\theta}\Bigg]=\dfrac{1}{2}\theta$$

$$\implies\,\tan^{-1}\Bigg[\dfrac{\tan\dfrac{\pi}{4}-\tan\theta}{1+\tan\dfrac{\pi}{4}\tan\theta}\Bigg]=\dfrac{1}{2}\theta$$

$$\implies\,\tan^{-1}\tan\bigg(\dfrac{\pi}{4}-\theta\bigg)=\dfrac{\theta}{2}$$

$$\implies\dfrac{\pi}{4}-\theta=\dfrac{\theta}{2}$$

$$\implies\theta=\dfrac{\pi}{6}=\tan^{-1}x$$

$$\implies x=tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$$
Question 10
1 / -0

If $$tan^{-1}x+2 cot^{-1}x=\dfrac{2\pi}{3}$$, then x is equal to

A
$$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

B
$$3$$

C
$$\sqrt{3}$$

D
$$\sqrt{2}$$

Solution

$$\tan^{-1}x+2 \cot^{-1}x=\dfrac{2\pi}{3}$$
$$\implies tan^{-1}x=2\bigg(\dfrac{\pi}{3}-\cot^{-1}x\bigg)=2\bigg(\dfrac{\pi}{3}-(\dfrac{\pi}{2}-\tan^{-1}x)\bigg)=2\bigg(-\dfrac{\pi}{6}+\tan^{-1}x\bigg)$$
$$\implies\tan^{-1}x=\dfrac{\pi}{3} \implies x=\tan\dfrac{\pi}{3}=\sqrt{3}$$

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Inverse Trigonometric Functions Test - 46

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Inverse Trigonometric Functions Test - 46

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Question 1

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

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

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

none of these

Solution

Question 2

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

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

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

none of these

Solution

Question 3

$$\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$$

$$\cos^{-1}\Bigg[\dfrac{b\cos\theta+a}{a\cos\theta+b}\Bigg]$$

$$\cos^{-1}\Bigg[\dfrac{a\cos\theta}{b\cos\theta+a}\Bigg]$$

$$\cos^{-1}\Bigg[\dfrac{b\cos\theta}{a\cos\theta+b}\Bigg]$$

Solution

Question 4

$$\sin^{-1}\bigg(\dfrac{1}{2}\bigg)-sin^{-1}x$$

$$\sin^{-1}x-\dfrac{\pi}{6}$$

$$\sin^{-1}x+\dfrac{\pi}{6}$$

none of these

Solution

Question 5

$$\dfrac{7}{13}$$

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

$$13$$

$$\dfrac{13}{7}$$

Solution

Question 6

$$\dfrac{1}{\sqrt{3}}$$

$$-\dfrac{1}{\sqrt{3}}$$

$${\sqrt{3}}$$

$$-\dfrac{\sqrt{3}}{4}$$

$$\dfrac{\sqrt{3}}{2}$$

Solution

Question 7

$$\dfrac{x}{2}$$

$$2x$$

$$3x$$

$$x$$

Solution

Question 8

$$0$$

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

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

none of these

Solution

Question 9

$$1$$

$$\sqrt{3}$$

$$\dfrac{1}{\sqrt{3}}$$

none of these

Solution

Question 10

$$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

$$3$$

$$\sqrt{3}$$

$$\sqrt{2}$$

Solution

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