Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

$$\sin \left\{\dfrac {\pi}{3}-\sin^{-1}\left (\dfrac {-1}{2}\right) \right\}=$$ ?

A
$$1$$

B
$$0$$

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

D
$$none\ of\ these$$

Solution
Question 2
1 / -0

Evaluate : $$\tan \dfrac {1}{2}\left (\cos^{-1}\dfrac {\sqrt 5}{3}\right)$$

A
$$\dfrac {(3-\sqrt 5)}{2}$$

B
$$\dfrac {(3+\sqrt 5)}{2}$$

C
$$\dfrac {(5-\sqrt 3)}{2}$$

D
$$\dfrac {(5+\sqrt 3)}{2}$$

Solution
Question 3
1 / -0

Evaluate : $$\cot (\tan^{-1}x+\cot^{-1}x)$$

A
$$1$$

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

C
$$0$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

If $$x\neq 0$$ then $$\cos (\tan^{-1}x+\cot^{-1}x)=$$?

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$none\ of\ these$$

Solution
Question 5
1 / -0

$$\cos \left(\tan^{-1} \dfrac {3}{4}\right)=$$?

A
$$\dfrac {3}{5}$$

B
$$\dfrac {4}{5}$$

C
$$\dfrac {4}{9}$$

D
$$none\ of\ these$$

Solution

Let $$\tan^{-1}\dfrac {3}{4}=x$$, where $$x\in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$$.

$$\therefore \ \tan x =\dfrac {3}{4}$$ and since $$x \in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$$, where have $$\cos x > 0$$

$$\therefore \ \cos x =\dfrac {1}{\sec x}=\dfrac {1}{\sqrt {1+\tan^2 x}}=\dfrac {1}{\sqrt {1+\dfrac {9}{16}}}=\dfrac {4}{5}$$

$$\therefore \ \cos \left (\tan^{-1}\dfrac {3}{4}\right)=\cos x =\dfrac {4}{5}$$.
Question 6
1 / -0

The value of $$\sin \left(\cos^{-1}\dfrac {3}{5}\right)$$ is

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

B
$$\dfrac {4}{5}$$

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

D
$$none\ of\ these$$

Solution
Question 7
1 / -0

Evaluate : $$\cos \left(2\tan^{-1}\dfrac {1}{2}\right)$$

A
$$\dfrac {3}{5}$$

B
$$\dfrac {4}{5}$$

C
$$\dfrac {7}{8}$$

D
$$none\ of\ these$$

Solution
Question 8
1 / -0

$$\sin \left [2\tan^{-1}\dfrac {5}{8}\right]$$

A
$$\dfrac {25}{64}$$

B
$$\dfrac {80}{89}$$

C
$$\dfrac {75}{128}$$

D
$$none\ of\ these$$

Solution
Question 9
1 / -0

$$\sin \left [2\sin^{-1}\dfrac{4}{5}\right]$$

A
$$\dfrac {12}{25}$$

B
$$\dfrac {16}{25}$$

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

D
$$none\ of\ these$$

Solution
Question 10
1 / -0

$$\tan \left\{\cos^{-1}\dfrac {4}{5}+\tan^{-1}\dfrac {2}{3}\right\}=$$ ?

A
$$\dfrac {13}{6}$$

B
$$\dfrac {17}{6}$$

C
$$\dfrac {19}{6}$$

D
$$\dfrac {23}{6}$$

Solution

$$\cos ^{ -1 }{ x } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } } \Rightarrow \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-\dfrac { 16 }{ 25 } } }{ \left( \dfrac { 4 }{ 5 } \right) } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } } $$

$$ \therefore \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \left( \dfrac { \dfrac { 3 }{ 4 } +\dfrac { 2 }{ 3 } }{ 1-\dfrac { 3 }{ 4 } \times \dfrac { 2 }{ 3 } } \right) } =\tan ^{ -1 }{ \dfrac { 17 }{ 6 } } $$

$$ \therefore$$ given exp. $$=\tan { \left\{ \tan ^{ -1 }{ \dfrac { 17 }{ 6 } } \right\} } =\dfrac { 17 }{ 6 } $$