Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

If $$\theta ={ cot }^{ -1 }\sqrt { cosx } -{ tan }^{ -1 }\sqrt { cosx } $$, then $$sin\theta =$$

A
$$tan\cfrac { 1 }{ 2 } x$$

B
$${ tan }^{ 2 }(x/2)$$

C
$$\cfrac { 1 }{ 2 } { tan }^{ -1 }(x2)$$

D
None of these

Solution
Question 2
1 / -0

The smallest and largest value of $$\tan^{-1}\left(\dfrac {1-x}{1+x}\right),0 \le x \le 1$$ are

A
$$0, \pi$$

B
$$0,\dfrac {\pi}{4}$$

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

D
$$\dfrac {\pi}{4}, \dfrac {\pi}{2}$$

Solution
Question 3
1 / -0

$$ \sin ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)+\sin ^{-1}\left(\dfrac{\sqrt{2}-1}{\sqrt{6}}\right)+\ldots+\sin ^{-1}\left(\dfrac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+\ldots . \infty= $$

A
$$\pi$$

B
$$\cfrac {\pi}{2}$$

C
$$\cfrac {\pi}{4}$$

D
$$\cfrac {3\pi}{2}$$

Solution
Question 4
1 / -0

If $$\cot \dfrac { 2{ x } }{ 3 } +\tan \dfrac { { x } }{ 3 } =\csc \dfrac { { kx } }{ 3 } ,$$ then the value of $$\tan ^{ { -1 } } (\tan { k } )$$ equals

A
$$2$$

B
$$2 - \pi$$

C
$$\pi - 2$$

D
$$2\pi - 2$$

Solution
Question 5
1 / -0

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

A
$$2-\sqrt{5}$$

B
$$\sqrt{5}-2$$

C
$$\frac{\sqrt{5}-2}{2}$$

D
5-$$\sqrt{2}$$

Solution
Question 6
1 / -0

$$4\tan ^{ -1 }{ \frac { 1 }{ 5 } } -\tan ^{ -1 }{ \frac { 1 }{ 70 } } +\tan ^{ -1 }{ \frac { 1 }{ 99 } } =$$

A
$$\pi $$

B
$${ \pi }/{ 2 }$$

C
$${ \pi }/{ 4 }$$

D
$${ 3\pi }/4$$
Question 7
1 / -0

Considering only the principal values of inverse functions, the set $$A=\left\{ x\quad \ge \quad 0\quad :tan^{ -1 }(2x)+tan^{ -1 }(3x)=\dfrac { \pi }{ 4 } \right\} $$

A
Is an empty set

B
contains more than two elements

C
contains two elements

D
is a singleton
Question 8
1 / -0

The value of $$\sin ^{ -1 }{ (\cos { (\cos ^{ -1 }{ (\cos { x } ) } +\sin ^{ -1 }{ (\sin { x } ) } ) } ) } ,\quad where\quad x\in (\frac { \pi }{ 2 } ,\pi )$$, is equal to

A
$$\frac { \pi }{ 2 } $$

B
$$-\pi $$

C
$$\pi $$

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

Solution
Question 9
1 / -0

The value of $$\sin^{-1}(\sin 12)-\cos^{-1}(\cos 12)=$$

A
$$0$$

B
$$\pi$$

C
$$8\pi +24$$

D
$$8\pi -24$$

Solution

$${\sin}^{-1}{\left(\sin{12}\right)}-{\cos}^{-1}{\left(\cos{12}\right)}$$
$$={\cos}^{-1}{\sqrt{1-{\sin}^{2}{12}}}-{\cos}^{-1}{\left(\cos{12}\right)}$$ since $${\sin}^{-1}{x}={\cos}^{-1}{\sqrt{1-{x}^{2}}}$$
$$={\cos}^{-1}{\left(\cos{12}\right)}-{\cos}^{-1}{\left(\cos{12}\right)}=0$$
Question 10
1 / -0

$$\tan ^{ -1 }{ (\frac { 5 }{ 12 } ) } +\sin ^{ -1 }{ (\frac { 24 }{ 25 } ) } =\cos ^{ -1 }{ (x) } \Rightarrow x=$$

A
$$\frac { -31 }{ 325 } $$

B
$$\frac { -33 }{ 325 } $$

C
$$\frac { -36 }{ 325 } $$

D
$$\frac { -39 }{ 325 } $$