<s>[INST] Inverse Trigonometry Question (Stuck in Algebraic Simplification) The original question: 

If x and y are of same sign, then find the value of $$\cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) $$

This is my attempt to the question:
$$\cfrac{x^3}{2\sin^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right)} + \cfrac{y^3}{2\cos^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right)} $$
$$
=\cfrac{x^3}{1-\cos(\tan^{-1} (x/y))} + \cfrac{y^3}{1+\cos(\tan^{-1}(y/x))} $$
$$
=\cfrac{x^3}{1 - \cfrac{|y|}{\sqrt{x^2+y^2}}} + \cfrac{y^3}{1+\cfrac{|x|}{\sqrt{x^2+y^2}}} $$
I'm not sure how to simplify this further. I've been trying to simplify this further for a long time but couldn't get any desirable answer.
Options are:
(A) $(x-y)(x^2+y^2)$
(B) $(x+y)(x^2-y^2)$
(C) $(x+y)(x^2+y^2)$
Also, any other better/alternate way to approach the question is welcomed. 
 [/INST] i will take $x,y$ to have the same sign. using double angle formula, we have  $$\tan^{-1}(y/x) = 2t, \tan^{-1}(x/y) = \pi/2 - 2t, r = \sqrt{x^2 + y^2}.$$ then 
$$\begin{align} \cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) &= \frac{x^3}{2\sin^2(\pi/4 - t)} + \frac{y^3}{2\cos^2 t}\\
&=\frac{x^3}{1 - \cos (\pi/2 - 2t)} + \frac{y^3}{1 + \cos 2t }\\
&=\frac{x^3}{1 - \sin 2t} + \frac{y^3}{1 + \cos 2t }\\
&=\frac{x^3}{1 - \frac y r} + \frac{y^3}{1 + \frac xr }\\
&= \frac{r\left(x^3(r+x)+y^3(r-y)\right)}{(r-y)(r+x)}\\
&=\frac{r\left( r(x^3+y^3)+(x^4 - y^4)\right)}{(r-y)(r+x)} \\
&=\frac{r(x+y)\left( r(x^2-xy+y^2)+(x - y)(x^2+y^2)\right)}{(r-y)(r+x)}\\
&=\frac{r^2(x+y)\left( r^2-xy +r(x - y)\right)}{(r-y)(r+x)}\\ 
&=r^2(x+y)=(x^2+y^2)(x+y) \end{align}$$
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