Browse Questions

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

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

Toolbox:
• $\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 $\large \frac{x+y}{1-xy}$
$tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg]$ = $\large \frac{x+y}{1-xy}$
edited Mar 14, 2013