Tan12 Sin−12x1+x2−12 Cos−11−y21+x2 =

Let x=Tan A, y=Tan BThen tan12sin-12x1+x2-12cos-11-y21+y2=tan12sin-12tanA1+tanA2-12cos-11-tanB21+tanB2 =tan12sin-1sin2A-12cos-1cos2B =tanA-B =tanA-tanB1+tanAtanB =x-y1+xy

Tan12 Sin−12x1+x2−12 Cos−11−y21+x2 =

Let x=Tan A, y=Tan BThen tan12sin-12x1+x2-12cos-11-y21+y2=tan12sin-12tanA1+tanA2-12cos-11-tanB21+tanB2 =tan12sin-1sin2A-12cos-1cos2B =tanA-B =tanA-tanB1+tanAtanB =x-y1+xy

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

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

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

answer is C.

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Detailed Solution

$L e t$ $x = T a n A,$ $y = T a n B$

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