Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

$$\tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$$, then

A
$$xy+yz+zx-xyz=0$$

B
$$xy+yz+zx+xyz=0$$

C
$$xy+yz+zx+1=0$$

D
$$xy+yz+zx-1=0$$

Solution

$$Given\,\,that \tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$$
$$\implies \tan^{-1}\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]=\dfrac{\pi}{2}$$
For this denominator of$$\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]$$ must be zero.
hence $$xy+yz+zx-1=0$$
Question 2
1 / -0

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

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\sqrt{2}$$

Solution

The given equation can be written as
$$3\tan^{-1}(\dfrac{1}{2+\sqrt{3}})=\dfrac{1}{3}+\tan^{-1}\dfrac{1}{x}$$
$$\implies 3(15^{\circ})=\tan^{-1}\dfrac{\dfrac{1}{x}+\dfrac{1}{3}}{1-\dfrac{1}{x}\dfrac{1}{3}}\,\implies1=\dfrac{3+x}{3x-1}\implies x=2$$
Question 3
1 / -0

If $$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$$, then $$\sin x$$ is

A
$$tan^2\dfrac{\alpha}{2}$$

B
$$cot^2\dfrac{\alpha}{2}$$

C
$$\tan \alpha$$

D
$$cot\dfrac{\alpha}{2}$$

Solution

$$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$$

$$tan^{-1}(\dfrac{1}{\sqrt{\cos a}})- tan^{-1}(\sqrt{\cos a})=x$$

$$tan^{-1}\dfrac{\dfrac{1}{\sqrt{\cos a}}-\sqrt{\cos a}}{1+\dfrac{1}{\sqrt{\cos a}}\sqrt{\cos a}}=x$$

$$tan^{-1}\dfrac{1-\cos a}{2\sqrt{\cos a}}=x$$

$$\dfrac{1-\cos a}{2\sqrt{\cos a}}=\tan x$$

$$\dfrac{2\sqrt{\cos a}}{1-\cos a}=\cot x$$

$$\sin x=\dfrac{1-\cos a}{1+\cos a}=\dfrac{2\sin^2\dfrac{\alpha}{2}}{2\cos^2\dfrac{\alpha}{2}}$$=$$tan^2\dfrac{\alpha}{2}$$
Question 4
1 / -0

The value of $$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))$$ is equal to

A
$$-1$$

B
$$\sqrt{2}$$

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

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

Solution

$$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))=cos(tan^{-1}(sin(tan^{-1}\infty)))$$
$$=cos(tan^{-1}(sin(\dfrac{\pi}{2}))$$
$$=cos(tan^{-1}(1))$$
$$=cos(\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}$$
Question 5
1 / -0

The domain of the function $$\cos ^{-1}(2x-1)$$ is

A
$$[0,1]$$

B
$$[-1,1]$$

C
$$(-1,1)$$

D
$$[0,\pi]$$

Solution

we have, $$f(x)=\cos ^{-1}(2x-1)$$
$$\because -1 \le 2x -1 \le1$$
$$\Rightarrow 0 \le 2x \le 2$$
$$\Rightarrow 0 \le x \le 1$$
$$\therefore x \in [0,1]$$
Question 6
1 / -0

If $$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$$ then $$x$$ is equal to

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

C
$$0$$

D
$$1$$

Solution

we have $$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$$
$$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \cos^{-1}0$$
$$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \cos^{-1} \cos \dfrac{\pi}{2}$$
$$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \dfrac{\pi}{2}$$
$$\Rightarrow \cos ^{-1}x= \dfrac{\pi}{2} - \sin ^{-1}\dfrac{2}{5}$$
$$\Rightarrow \cos ^{-1}x=\cos ^{-1} \dfrac{2}{5} \ \ [\because \cos^{-1}x+ \sin ^{-1}x=\dfrac{\pi}{2}]$$
$$\therefore x=\dfrac{2}{5}$$
Question 7
1 / -0

The value of $$\sin [2 \tan ^{-1} (.75)]$$ is equal to

A
$$0.75$$

B
$$1.5$$

C
$$0.96$$

D
$$\sin 1.5$$

Solution

we have, $$\sin [2 \tan ^{-1} (.75)] = \sin \left(2 \tan ^{-1}\dfrac{3}{4}\right) [0.75=\dfrac{75}{100}=\dfrac{3}{4}$$
$$=\sin \left(\sin ^{-1}\dfrac{2.\dfrac{3}{4}}{1+\dfrac{9}{16}}\right)=\sin \left[\sin ^{-1} \dfrac{3/2}{25/16}\right]$$
$$=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\dfrac{24}{25}=0.96$$
Question 8
1 / -0

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

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

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

C
$$\dfrac{5\pi}{2}$$

D
$$\dfrac{7\pi}{2}$$

Solution

We have, $$\cos ^{-1} \left(\cos \dfrac{3 \pi}{2}\right)$$
$$=\cos ^{-1}\cos \left( 2 \pi \dfrac{ \pi}{2}\right) \ \ \ \left[\because \cos \left(2 \pi -\dfrac{\pi}{2}\right)\right]= \cos \dfrac{\pi}{2}$$

$$=\cos ^{-1}\cos \left( \dfrac{ \pi}{2}\right) =\dfrac{ \pi}{2} \ \ \ \left\{\because \cos^{-1} (\cos x)=x, x \in [0,\pi] \right\}$$
Note remember that, $$\cos ^{-1} \left(\cos \dfrac{3 \pi}{2}\right) \ne \dfrac{3 \pi}{2}$$
$$\because \dfrac{3 \pi }{2} \in (0, \pi)$$
Question 9
1 / -0

If $$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$$, then $$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$$

A
$$0$$

B
$$1$$

C
$$6$$

D
$$12$$

Solution

we have,$$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$$
$$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$$

we know that , $$ 0 \le \cos^{-1} x \le \pi$$
$$\Rightarrow \cos^{-1} \alpha +\cos^{-1} \beta +\cos^{-1}
\gamma = 3 \pi$$
If and only if, $$\cos^{-1}\alpha=\cos^{-1} \beta =\cos^{-1}
\gamma =\pi$$
$$\Rightarrow \cos \pi=\alpha=\beta=\gamma$$
$$\Rightarrow -1=\alpha=\beta=\gamma$$
$$\therefore \alpha (\beta + \gamma) + \beta (\gamma +
\alpha)+ \gamma (\alpha + \beta)$$
$$=-1(-1-1)-1(1-1)-1(-1-1)$$$$=2+2+2=6$$
Question 10
1 / -0

The value of $$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$$ is

A
$$\dfrac{25}{24}$$

B
$$\dfrac{25}{7}$$

C
$$\dfrac {24} {25}$$

D
$$\dfrac {7} {24}$$

Solution

we have $$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$$
Let $$\cos ^{-1}\dfrac{7}{25}=x$$
$$\Rightarrow \cos x=\dfrac{7}{25}$$
$$\therefore \sin x=\sqrt{1-\cos^2x=}\sqrt{1-\left(\dfrac{7}
{25}\right)^2}$$
$$=\sqrt{\dfrac{625-49}{625}}=\dfrac{24}{25}$$
$$\therefore \cot x=\dfrac{\cos x}{\sin x}= \dfrac{\dfrac{7}
{25}}{\dfrac{24}{25}}=\dfrac{7}{24}$$

$$\Rightarrow x=\cot^{-1} \left(\dfrac{7}{24} \right)=\cot^{-1} \left(\dfrac{7}{25}\right)$$

$$\therefore \cot \left(\cos^{-1}\dfrac{7}{25}\right)= \cot
\left(\cos^{-1}\dfrac{7}{24}\right)=\dfrac{7}{24} \ \ \left
[\because \cos^{-1}\dfrac{7}{24}=\cos^{-1}\dfrac{7}
{25}\right]$$

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Inverse Trigonometric Functions Test - 47

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Inverse Trigonometric Functions Test - 47

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Question 1

$$xy+yz+zx-xyz=0$$

$$xy+yz+zx+xyz=0$$

$$xy+yz+zx+1=0$$

$$xy+yz+zx-1=0$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$\sqrt{2}$$

Solution

Question 3

$$tan^2\dfrac{\alpha}{2}$$

$$cot^2\dfrac{\alpha}{2}$$

$$\tan \alpha$$

$$cot\dfrac{\alpha}{2}$$

Solution

Question 4

$$-1$$

$$\sqrt{2}$$

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

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

Solution

Question 5

$$[0,1]$$

$$[-1,1]$$

$$(-1,1)$$

$$[0,\pi]$$

Solution

Question 6

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

$$0$$

$$1$$

Solution

Question 7

$$0.75$$

$$1.5$$

$$0.96$$

$$\sin 1.5$$

Solution

Question 8

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

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

$$\dfrac{5\pi}{2}$$

$$\dfrac{7\pi}{2}$$

Solution

Question 9

$$0$$

$$1$$

$$6$$

$$12$$

Solution

Question 10

$$\dfrac{25}{24}$$

$$\dfrac{25}{7}$$

$$\dfrac {24} {25}$$

$$\dfrac {7} {24}$$

Solution

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