First slide

Properties of ITF

Question

$\tan [\frac{1}{2} S i n^{- 1} (\frac{2 x}{1 + x^{2}}) - \frac{1}{2} C o s^{- 1} (\frac{1 - y^{2}}{1 + y^{2}})] =$

Easy

A

$\frac{2 x}{1 - x^{2}}$
B

0
C

$\frac{x - y}{1 + x y}$
D

1

Solution

$L e t x = \tan A; y = \tan B . T h e n \tan [\frac{1}{2} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) - \frac{1}{2} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] = \tan [\frac{1}{2} \sin^{- 1} (\frac{2 \tan A}{1 + \tan A^{2}}) - \frac{1}{2} \cos^{- 1} (\frac{1 - \tan B^{2}}{1 + \tan B^{2}})] = \tan [\frac{1}{2} \sin^{- 1} (\sin 2 A) - \frac{1}{2} \cos^{- 1} (\cos 2 B)] = \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{x - y}{1 + x y}$

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