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

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

## 1 Answer

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 $$\large \frac{x+y}{1-xy}$$
$$tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg]$$ = $$\large \frac{x+y}{1-xy}$$
answered Feb 22, 2013
edited Mar 14, 2013

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