First slide

Properties of ITF

Question

If $\sin^{- 1} (\frac{2 a}{1 + a^{2}}) + \sin^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 \tan^{- 1} x,$ then $x,$ is equal to

Moderate

A

$\frac{a - b}{1 + a b}$
B

$\frac{b}{1 + a b}$
C

$\frac{b}{1 - a b}$
D

$\frac{a + b}{1 - a b}$

Solution

We have,
$\sin^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 \tan^{- 1} x$
$\sin^{- 1} \frac{2 a}{1 + a^{2}} + \sin^{- 1} \frac{2 b}{1 + b^{2}} = 2 \tan^{- 1} x$
$\Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} b = 2 \tan^{- 1} x$
$\begin{array}{l} = \tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} x \\ = \tan^{- 1} (\frac{a + b}{1 - a b}) = \tan^{- 1} x \Rightarrow x = \frac{a + b}{1 - a b} \end{array}$

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