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Find the value of \[ tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg], | x | < 1, y > 0 \: and \: xy < 1\]

$\begin{array}{1 1} (\large \frac{x+y}{1-xy} ) \\ (\large \frac{x-y}{1+xy} ) \\ (\large \frac{1+xy}{x-y} ) \\ (\large \frac{1-xy}{x+y} ) \end{array} $

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  • \( \large \frac{2tan\theta}{1+tan^2\theta}\)\(=sin2\theta\)
  • \(\large \frac{1-tan^2\theta}{1+tan^2\theta}\)\(=cos2\theta\)
  • \( tan(A+B)=\large \frac{tanA+tanB}{1-tanA.tanB}\)
Given $tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg], | x | < 1, y > 0 \: and \: xy < 1$
Let \( x=tan \alpha \:,\: y=tan \beta\) \( \Rightarrow \alpha = tan^{-1}x\: ,\: \beta=tan^{-1}y\)
Given $ x=tan \alpha$ and \( \large \frac{2tan\theta}{1+tan^2\theta}\)\( = sin2\theta\)\(\rightarrow \large \frac{2x}{1+x^2}\)\(=\large \frac{2tan\alpha}{1+tan^2\alpha}\)\( = sin 2\alpha\)
Similarly, given $ y=tan \beta$ and \( \large \frac{1-tan^2\theta}{1+tan^2\theta}\)\( = cos2\theta\)\(\rightarrow \large \frac{1-y^2}{1+y^2}\)\(=\large \frac{1-tan^2\beta}{1+tan^2\beta}\)\( = cos 2\alpha\)
The given expression reduces to \(tan\frac{1}{2} \bigg[ sin^{-1}sin2\alpha+cos^{-1}cos2\beta \bigg] \)
Given that $sin^{-1} sin x = x$ and $cos^{-1}cos x = x$, this further reduces to \( tan\big(\frac{1}{2} (2\alpha+2\beta)\big)= tan(\alpha+\beta)\)
We know that \( tan(A+B)=\large \frac{tanA+tanB}{1-tanA.tanB}\), and substituting the value of tan\(\alpha=x\:,\:tan\beta=y\:\):
$tan(\alpha+\beta)$ reduces to \( \large \frac{x+y}{1-xy} \)
\( tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg]\) = \( \large \frac{x+y}{1-xy} \)
answered Feb 22, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai
 

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