$\tan \{\frac{1}{2} \sin^{–1} (2x/ 1 + x^2) + \frac{1}{2} \cos^{–1} (1 – y^2/1 + y^2) \}$ is equal to. $$
\tan \left\{
\frac{1}{2} \arcsin \frac{2x}{1 + x^2} +
\frac{1}{2} \arccos \frac{1 – y^2}{1 + y^2} 
\right\}
$$ 
is equal to.
Note:
i think $\sin a=2x/1+x^2$, $\cos b=(1 – y^2/1 + y^2)$

Set $\arctan x=A\implies x=\tan A\  \ \ \ (1)$
Using definition of Principal values, $\displaystyle-\frac\pi2\le A\le\frac\pi2\  \ \ \ (2)$
and using Weierstrass substitution formula, $\displaystyle\frac{2x}{1+x^2}=\sin2A$
Now $\displaystyle\arcsin(\sin2A)=\begin{cases} 2A &\mbox{if } -\dfrac\pi2\le2A\le\dfrac\pi2\iff-\dfrac\pi4\le A\le\dfrac\pi4 \\ 
\pi-2A & \text{otherwise} \end{cases} $
From $(1),(2); \cos A\ge0\implies\cos A=\dfrac1{\sec A}=+\dfrac1{\sqrt{1+\tan^2A}}=\dfrac1{\sqrt{1+x^2}}$
$\sin A=\cos A\cdot\tan A=\dfrac x{\sqrt{1+x^2}}$
Similarly,  setting $\arctan y=B,\tan B=y\  \ \ \ (3)$
and like $(2),\displaystyle-\frac\pi2\le B\le\frac\pi2$
$(3)\implies\dfrac{1-y^2}{1+y^2}=\cos2B$
$\displaystyle\arccos(\cos2B)=\begin{cases} 2B &\mbox{if } 0\le2B\le\pi\iff0\le B\le\dfrac\pi2 \\ 
-2B & \text{otherwise} \end{cases} $
Can you take care of the four possible combinations?
