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Inverse Trigonometric Functions

Multiple Choice Questions

121.

The value of $\sin [2 \cos^{- 1} (\frac{\sqrt{5}}{3})]$ is

$\frac{\sqrt{5}}{3}$
$\frac{2 \sqrt{5}}{3}$
$\frac{4 \sqrt{5}}{9}$
$\frac{2 \sqrt{5}}{9}$

122.

$\sin (2 \sin^{- 1} (\sqrt{\frac{63}{65}}))$ is equal to

$\frac{2 \sqrt{126}}{65}$
$\frac{4 \sqrt{65}}{65}$
$\frac{8 \sqrt{63}}{65}$
$\frac{\sqrt{63}}{65}$

123.
$\cot^{- 1} (2 . 1^{2}) + \cot^{- 1} (2 . 2^{2}) + \cot^{- 1} (2 . 3^{2}) + . . .$ upto $\infty$ is equal to

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{π}{5}$

$\frac{π}{4}$

$\cot^{- 1} (2 . 1^{2}) + \cot^{- 1} (2 . 2^{2}) + \cot^{- 1} (2 . 3^{2}) + . . . \infty = \sum_{r = 1}^{\infty} \cot^{- 1} (2 . r^{2}) = \sum_{r = 1}^{\infty} \tan^{- 1} (\frac{1}{2 r^{2}}) = \sum_{r = 1}^{\infty} \tan^{- 1} (\frac{(1 + 2 r) + (1 - 2 r)}{1 - (1 + 2 r) (1 - 2 r)}) = \sum_{r = 1}^{\infty} [\tan^{- 1} (1 + 2 r) + \tan^{- 1} (1 - 2 r)] = \tan^{- 1} (3) - \tan^{- 1} (1) + \tan^{- 1} (5) - \tan^{- 1} (3) + \tan^{- 1} (7) - \tan^{- 1} (5) + . . . + \tan^{- 1} (\infty) = - \frac{π}{4} + \frac{π}{2} = \frac{π}{4}$

124.

If 'x' takes negative permissible value, then sin^-1(x) is equal to

$- \cos^{- 1} (\sqrt{1 - x^{2}})$
$\cos^{- 1} (\sqrt{x^{2} - 1})$
$π - \cos^{- 1} (\sqrt{1 - x^{2}})$
$\cos^{- 1} (\sqrt{1 - x^{2}})$

125.

If a > b > 0, $\sec^{- 1} (\frac{a + b}{a - b}) = 2 \sin^{- 1} (x)$ , then x is

$- \sqrt{\frac{b}{a + b}}$
$\sqrt{\frac{b}{a + b}}$
$- \sqrt{\frac{a}{a + b}}$
$\sqrt{\frac{a}{a + b}}$

126.

If $x \neq nπ, x \neq (2 n + 1) \frac{π}{2}, n \in Z$ , then $\frac{\sin^{- 1} (\cos (x)) + \cos^{- 1} (\sin (x))}{\tan^{- 1} (\cot (x)) + \cot^{- 1} (\tan (x))}$ is