Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 25

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

If $$\theta\equiv sin^{-1}x+cos^{-1}x-tan^{-1}x,\ 0\leq x\leq 1$$, then the smallest interval in which $$\theta$$ lies is given by ?

A
$$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$$

B
$$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$$

C
$$0\displaystyle \leq\theta\leq\frac{\pi}{4}$$

D
$$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$$

Solution

Given $$0\le x\le 1\implies \tan^{-1} x\in [\tan^{-1}0,\tan^{-1}1]\implies \tan^{-1} x\in [0,\frac{\pi}{4}]\implies -\tan^{-1} x\in [-\frac{\pi}{4},0]$$

$$\theta=\sin^{-1} x+\cos^{-1} x-\tan^{-1} x=\dfrac{\pi}{2} -\tan^{-1} x$$ $$(\because \sin^{-1} x+\cos^{-1} x=\dfrac{\pi}{2})$$

As $$-\tan^{-1} x\in [-\frac{\pi}{4},0]$$

$$\implies \dfrac{\pi}{2}-\tan^{-1} x\in [\frac{\pi}{4},\frac{\pi}{2}]$$

$$\implies \dfrac{\pi}{4}\le\theta\le \dfrac{\pi}{2}$$
Question 2
1 / -0

The value of sin $$\sin \left\{ {{{\tan }^{ - 1}}\left( {\tan \frac{{7\pi }}{6}}

\right) + {{\cos }^{ - 1}}\left( {\cos \frac{{7\pi }}{3}} \right)}

\right\}$$ is

A
0

B
1

C
-1

D
None of these

Solution

$$Sin \left \{ \underbrace{tan^-1 \left (tan \frac{7\pi}{6} \right )} + \underbrace{cos^{-1}\left ( cos \frac{7\pi}{3} \right )} \right \}$$
$$A$$ $$B$$
$$A=tan^{-1} \left ( tan \left ( \pi + \frac{\pi}{6} \right ) \right )$$
$$=tan^{-1} \left ( tan \frac{\pi}{6} \right ) =\frac{\pi}{6}$$
$$B=cos^{-1} \left ( cos \left ( 2\pi + \frac{\pi}{3} \right ) \right )$$
$$=\frac{\pi}{3}$$ $$\therefore sin \left ( \frac{\pi}{6}+ \frac{\pi}{3} \right ) = sin \left ( \frac{\pi}{2} \right ) =1$$
Question 3
1 / -0

If $$\alpha \epsilon \left ( 0,\dfrac{\pi}{2}\right )$$, then the value of $$\tan ^{-1}(\cot \alpha )-\cot ^{-1}(\tan \alpha )+\sin ^{-1}(\sin \alpha )-\cos ^{-1}(\cos \alpha )$$ is equal to

A
$$2\alpha $$

B
$$\pi +\alpha $$

C
$$0$$

D
$$\pi -2\alpha$$

Solution

$$tan^{ -1 }\left( cot\alpha \right) -cot^{ -1 }\left( tan\alpha \right) +sin^{ -1 }\left( sin\alpha \right) -cos^{ -1 }\left( cos\alpha \right) \\ =tan^{ -1 }\left( tan\left( \cfrac { \pi }{ 2 } -\alpha \right) \right) -cot^{ -1 }\left( cot\left( \cfrac { \pi }{ 2 } -\alpha \right) \right) +sin^{ -1 }sin\alpha -cos^{ -1 }cos\alpha \\ =\left( \cfrac { \pi }{ 2 } -\alpha \right) -\left( \cfrac { \pi }{ 2 } -\alpha \right) +\alpha -\alpha =0$$
Question 4
1 / -0

If $$\tan ^{-1}\dfrac{\sqrt{1+x^{2}}-1}{x}=4^{\circ}$$,then

A
$$x=\tan 2^{\circ}$$

B
$$x=\tan 4^{\circ}$$

C
$$x=\tan (1/4)^{\circ}$$

D
$$x=\tan 8^{\circ}$$

Solution

Substituting $$x=tanA$$

$$\tan^{-1}(\dfrac{\sec A-1}{\tan A})$$

           $$=\tan^{-1}(\dfrac{1-\cos A}{\sin A})$$

                 $$=\tan^{-1}(\tan(\dfrac{A}{2}))$$............Using $$\cos A=1-2\sin^2\dfrac A2$$

$$4^{0}$$     $$=\dfrac{A}{2}$$

$$\Rightarrow A=8^{0}$$

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

$$x=\tan(8^{0})$$
Question 5
1 / -0

If $${\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}$$ then $$x+y+z$$ equals

A
$$xyz$$

B
$$xy+yz+zx$$

C
$$2xyz$$

D
None of these

Solution

$$\cot^{-1}x+\cot^{-1}y=\dfrac{\pi}{2} -\cot^{-1}z$$
$$\tan(\cot^{-1}x+\cot^{-1}y)=\tan \left ( \dfrac{\pi}{2} - \cot^{-1}z \right )$$
$$\displaystyle \dfrac{\tan(\cot^{-1}x)+\tan (\cot^{-1}y)}{1- \tan (\cot^{-1}x)\tan (\cot^{-1}y)}=z$$
$$\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{1-\dfrac{1}{xy}}=z$$
$$\dfrac{x+y}{xy-1}=z$$
$$x+y+z=xyz$$
Question 6
1 / -0

If $$sin\left\{ \sin ^{ -1 }{ \cfrac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right\} =1$$, then $$x$$ is equal to

A
$$1$$

B
$$0$$

C
$$\displaystyle\frac{ 4 }{ 5 }$$

D
$$\displaystyle\frac{ 1 }{ 5 }$$

Solution

Given, $$sin \left(sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x) \right)=1$$

$$sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x)=sin^{-1}(1)$$

$$sin^{-1} \left(\displaystyle\frac{1}{5} \right)+cos^{-1}(x)=\dfrac{\pi}{2}$$

$$cos^{-1}(x)=\displaystyle\frac{\pi}{2}-sin^{-1} \left(\displaystyle\frac{1}{5} \right)$$

$$cos^{-1}(x)=cos^{-1} \left(\displaystyle\frac{1}{5} \right)$$

$$\therefore x=\displaystyle\frac{1}{5}$$
Question 7
1 / -0

If $$0\le x\le 1$$, then $$\tan { \left\{ \cfrac { 1 }{ 2 } \sin ^{ -1 }{ \cfrac { 2x }{ 1+{ x }^{ 2 } } +\cfrac { 1 }{ 2 } \cos ^{ -1 }{ \cfrac { 2x }{ 1+{ x }^{ 2 } } } } \right\} } $$

A
1

B
$$0$$

C
$$\cfrac { 2x }{ 1+{ x }^{ 2 } } $$

D
$$x$$

Solution

$$tan[\frac{1}{2}(sin^{-1}(\frac{2x}{1+x^2})+cos^{-1}(\frac{2x}{1+x^2}))]$$
$$=tan[\frac{1}{2}(sin^{-1}(\theta)+cos^{-1}(\theta))]$$
$$=tan(\frac{\pi}{2(2)})$$
$$=tan(\frac{\pi}{4})$$
$$=1$$
Question 8
1 / -0

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

A
$$\cfrac { \pi }{ 2 } -x$$

B
$$\cot { \left( \tan ^{ -1 }{ x } \right) } $$

C
$$\tan { x } $$

D
none of these

Solution

Given, $$\tan(cot^{-1}(x))$$
$$=\tan(\tan^{-1}(\dfrac{1}{x}))$$
$$=\dfrac{1}{x}$$.
Also
$$\tan(\cot^{-1}(x))$$
$$=\tan(\dfrac{\pi}{2}-\tan^{-1}(x))$$
$$=\cot(\tan^{-1}(x))$$
Question 9
1 / -0

If $$\displaystyle x$$ takes negative permissible value, then $$\displaystyle \sin ^{-1}x$$ is equal to

A
$$\displaystyle \cos ^{-1}\sqrt{1-x^{2}}$$

B
$$\displaystyle -\cos ^{-1}\sqrt{1-x^{2}}$$

C
$$\displaystyle \cos ^{-1}\sqrt{x^{2}-1}$$

D
$$\displaystyle \pi -\cos ^{-1}\sqrt{1-x^{2}}$$

Solution

$$sin^{-1}(x)$$
$$=cos^{-1}(\sqrt{1-x^2})$$ , for $$x>0$$
Since x takes negative permissible value.
$$sin^{-1}(x)=-cos^{-1}(\sqrt{1-x^2})$$
Question 10
1 / -0

$$\sec ^{ 2 }{ \left( \tan ^{ -1 }{ 2 } \right) +cosec ^{ 2 }{ \left( \cot ^{ -1 }{ 3 } \right) } } $$=

A
5

B
10

C
15

D
20

Solution

The given question can be re-written as
$$1+\tan^{2}(\tan^{-1}(2))+1+\cot^{2}(\cot^{-3})$$
$$=2+2^{2}+3^{2}$$
$$=2+4+9$$
$$=15$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 25

Result

Inverse Trigonometric Functions Test - 25

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$$

$$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$$

$$0\displaystyle \leq\theta\leq\frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$$

Solution

Question 2

0

1

-1

None of these

Solution

Question 3

$$2\alpha $$

$$\pi +\alpha $$

$$0$$

$$\pi -2\alpha$$

Solution

Question 4

$$x=\tan 2^{\circ}$$

$$x=\tan 4^{\circ}$$

$$x=\tan (1/4)^{\circ}$$

$$x=\tan 8^{\circ}$$

Solution

Question 5

$$xyz$$

$$xy+yz+zx$$

$$2xyz$$

None of these

Solution

Question 6

$$1$$

$$0$$

$$\displaystyle\frac{ 4 }{ 5 }$$

$$\displaystyle\frac{ 1 }{ 5 }$$

Solution

Question 7

1

$$0$$

$$\cfrac { 2x }{ 1+{ x }^{ 2 } } $$

$$x$$

Solution

Question 8

$$\cfrac { \pi }{ 2 } -x$$

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

$$\tan { x } $$

none of these

Solution

Question 9

$$\displaystyle \cos ^{-1}\sqrt{1-x^{2}}$$

$$\displaystyle -\cos ^{-1}\sqrt{1-x^{2}}$$

$$\displaystyle \cos ^{-1}\sqrt{x^{2}-1}$$

$$\displaystyle \pi -\cos ^{-1}\sqrt{1-x^{2}}$$

Solution

Question 10

5

10

15

20

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.