Question

Find the value of $tan\frac{1}{2}[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}]$
$|x|<1,y>0andxy<1$

• A. $\frac{x+y}{1-xy}$
• B. $\pi +\frac{x+y}{1-xy}$
• C. $\frac{1-xy}{x+y}$
• D. $\pi +\frac{1-xy}{x+y}$

Answer 1

The correct option is A $\frac{x+y}{1-xy}$

Let $x=tan\theta$, then $\theta =ta{n}^{-1}x$.
$\therefore si{n}^{-1}\frac{2x}{1+x^2}=si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$
$=si{n}^{-1}\left(sin2\theta \right)=2\theta =2ta{n}^{-1}x$
Let $y=tan\varphi ,Then\varphi =ta{n}^{-1}y$
$\therefore co{s}^{-1}\frac{1-y^2}{1+y^2}=co{s}^{-1}\left(\frac{1-ta{n}^2\varphi }{1+ta{n}^2\varphi }\right)$
$=co{s}^{-1}\left(cos2\varphi \right)=2\varphi =2ta{n}^{-1}y$
$\therefore tan\frac{1}{2}[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}]$
$=tan\frac{1}{2}[2ta{n}^{-1}x+2ta{n}^{-1}y]$
$=tan[ta{n}^{-1}x+ta{n}^{-1}y]$
$=tan\left[ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)\right]=\frac{x+y}{1-xy}$