Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Multiple choice Questions :
If a > b > c ,
$$ \cot^{-1} \left ( \dfrac{1 + ab}{a - b} \right ) + \cot^{-1}\left ( \dfrac{1 + bc}{b - c} \right ) + \cot^{-1} \left ( \dfrac{1 + ac}{c - a} \right ) $$

A
$$ 0 $$

B
$$ \pi $$

C
$$ 2 \pi $$

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

Solution

$$ \tan ^{-1}\left ( \dfrac{a - b}{1 + ab} \right ) + \tan^{-1} \left ( \dfrac{b - c}{1 + bc} \right ) + \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) $$
$$ = \tan^{-1} a - \tan^{-1} b + \tan^{-1} b - \tan^{-1} c + \pi - \tan^{-1} a + \tan^{-1} c $$
$$ = \pi \ \ \ \ \ \ \ \ as \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) = \pi - \tan^{-1} \left ( \dfrac{a - c}{1 + ac} \right )$$
Question 2
1 / -0

Multiple choice Questions :
$$ \sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )=$$

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

B
$$ 0 $$

C
$$ \pi $$

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

Solution

$$ \sin ^{1} x + \cos^{-1} x= \dfrac{\pi}{2} $$
$$ \sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )= \dfrac{\pi}{2} $$
Question 3
1 / -0

Multiple choice Questions :
$$ \cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = $$

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

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

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

D
$$ - \pi $$

Solution

$$ \cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = \cos^{1}\cos \left ( \pi + \dfrac{\pi}{3} \right ) $$
$$ = \cos^{-1} \left [ - cos \dfrac{\pi}{3} \right ] $$
$$ = \pi - cos^{-1} \left ( \cos\dfrac{\pi}{3} \right ) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$$
Question 4
1 / -0

$$\tan ^{-1}\left ( \dfrac{x}{y} \right ) - \tan^{-1} \dfrac{x - y}{x + y}$$ is equal to

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

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

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

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

Solution

$$\tan^{-1} \left ( \dfrac{x}{y} \right ) - \left ( \tan^{-1}\dfrac{\frac{x}{y}-1}{1+\dfrac{x}{y}} \right )$$
$$=\tan^{-1} \left ( \dfrac{x}{y} \right )-\tan ^{-1} \left ( \dfrac{x}{y} \right )+ \tan^{-1} (1)$$
$$\tan^{-1} 1 = \dfrac{\pi }{4}$$
Question 5
1 / -0

$$2\tan(\tan^{-1}x+\tan^{-1}x^3)$$ is:

A
$$\dfrac{2x}{1-x^2}$$

B
$$1+x^2$$

C
$$2x$$

D
none of these

Solution

$$2\tan (\tan^{-1}x+\tan^{-1}x^{3})$$
$$=2\tan \left\{\tan^{-1}\left(\dfrac{x+x^{3}}{1-x\times x^{3}}\right)\right\}$$
$$=2\tan \left\{\tan^{-1}\left(\dfrac{x(1+x^{2})}{1-x^{4}}\right)\right\}$$
$$=2\tan \left\{\tan^{-1}\dfrac{x(1+x^{2})}{(1-x^{2})(1+x^{2})}\right\}$$
$$=2\tan \dfrac{x}{1-x^{2}}=\dfrac{2x}{1-x^{2}}$$
Hence, option $$(a)$$ is correct.
Question 6
1 / -0

If $$\sin^{-1}\left(\dfrac{1}{2}\right)=x$$, then general value $x$$ is:

A
$$2n\pi \pm \dfrac{\pi}{6}$$

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

C
$$n\pi \pm \dfrac{\pi}{6}$$

D
$$n\pi +(-1)^n\dfrac{\pi}{6}$$

Solution

Given,
$$\sin^{-1}\left(\dfrac{1}{2}\right)=x$$
$$\Rightarrow \sin x=\dfrac{1}{2}=\sin \dfrac{\pi}{6}$$
$$\Rightarrow x=\dfrac{\pi}{6}$$
$$\therefore$$ General value of $$x, \theta=n\pi +(-1)^{n}\dfrac{\pi}{6}$$
Hence, option $$(d)$$ is correct.
Question 7
1 / -0

If $$\tan^{-1}(3x)+\tan^{-1} 2x=\dfrac{\pi}{4}$$, then $$x$$ is:

A
$$\dfrac{1}{6}$$

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

C
$$\dfrac{1}{10}$$

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

Solution

$$\tan^{-1}(3x)+\tan^{-1}(2x)=\dfrac{\pi}{4}$$
$$\Rightarrow \tan^{-1}\left\{\dfrac{3x+2x}{1-3x\times 2x}\right\}=\dfrac{\pi}{4}$$
$$\Rightarrow \left(\dfrac{5x}{1-6x^{2}}\right)=\tan \dfrac{\pi}{4}$$
$$\Rightarrow \dfrac{5x}{1-6x^{2}}=1$$
$$\Rightarrow 1-6x^{2}=5x$$
$$\Rightarrow 6x^{2}+5x-1=0$$
$$\Rightarrow 6x^{2}+6x-x-1=0$$
$$\Rightarrow 6x(x+1)-1(x+1)=0$$
$$\Rightarrow (x+1)(6x-1)=0$$
$$\Rightarrow x=-1, x=\dfrac{1}{6}$$
Hence, option $$(a)$$ is correct.
Question 8
1 / -0

If $$\tan^{-1}(1)+\cos^{-1}(\dfrac{1}{\sqrt{2}})=\sin^{-1}x$$, then value of $$x$$ is

A
$$-1$$

B
$$0$$

C
$$1$$

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

Solution

$$\tan^{-1}(1)+\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)=\sin^{-1}x$$
$$\Rightarrow \sin^{-1}x=\dfrac{\pi}{4}+\dfrac{\pi}{4}$$
$$\Rightarrow \sin^{-1}x=\dfrac{\pi}{2}$$
$$\Rightarrow x=\sin \dfrac{\pi}{2}$$
$$\Rightarrow x=1$$
Hence, option $$(c)$$ is correct.
Question 9
1 / -0

Value of $$\sin^{-1}(\dfrac{\sqrt{3}}{2})+2\cos^{-1}(\dfrac{\sqrt{3}}{2})$$ is:

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

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

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

D
$$\pi$$

Solution

$$\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+2\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$
$$=\left[\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right]+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$
$$=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2\pi}{3}\left[\because \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\right]$$
Hence, option $$(c)$$ is correct.
Question 10
1 / -0

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

A
$$1$$

B
$$3$$

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

D
none of these

Solution

$$\cot^{-1}(x)+\tan^{-1}\left(\dfrac{1}{3}\right)=\dfrac{\pi}{2}$$
$$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{3}\right)$$
$$\Rightarrow \cot^{-1}x=\cot^{-1}\dfrac{1}{3}$$
On comparing $$x=\dfrac{1}{3}$$
Hence, option $$(c)$$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 50

Result

Inverse Trigonometric Functions Test - 50

Score

-

Rank

-

SHARING IS CARING

Question 1

$$ 0 $$

$$ \pi $$

$$ 2 \pi $$

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

Solution

Question 2

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

$$ 0 $$

$$ \pi $$

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

Solution

Question 3

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

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

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

$$ - \pi $$

Solution

Question 4

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

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

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

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

Solution

Question 5

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

$$1+x^2$$

$$2x$$

none of these

Solution

Question 6

$$2n\pi \pm \dfrac{\pi}{6}$$

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

$$n\pi \pm \dfrac{\pi}{6}$$

$$n\pi +(-1)^n\dfrac{\pi}{6}$$

Solution

Question 7

$$\dfrac{1}{6}$$

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

$$\dfrac{1}{10}$$

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

Solution

Question 8

$$-1$$

$$0$$

$$1$$

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

Solution

Question 9

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

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

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

$$\pi$$

Solution

Question 10

$$1$$

$$3$$

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

none of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.