Write the correct alternative in the following:

Given:

$y=\frac{2}{\sqrt{\left(a^{2}-b^{2}\right)}} \tan ^{-1}\left(\frac{a-b}{a+b} \tan \frac{x}{2}\right)$

$\frac{d y}{d x}=\frac{2}{\sqrt{\left(a^{2}-b^{2}\right)}}\left(\frac{1}{1+\left(\frac{a-b}{a+b} \tan \frac{x}{2}\right)^{2}}\right)\left(\frac{a-b}{a+b}\right)\left(\sec \frac{x}{2}\right)^{2}$

$=\frac{2}{\sqrt{\left(a^{2}-b^{2}\right)}}\left(\frac{(a+b)^{2}}{(a+b)^{2}+(a-b)^{2}\left(\tan \frac{x}{2}\right)^{2}}\right)\left(\frac{a-b}{a+b}\right)\left(\sec \frac{x}{2}\right)^{2}$

$=\frac{2}{\left.\sqrt{(} a^{2}-b^{2}\right)}\left(\frac{(a+b)}{a^{2}\left(1+(\tan x)^{2}\right)+b^{2}\left(1+(\tan x)^{2}\right)+2 a b\left(1-(\tan x)^{2}\right)}\right)(a$

$-b)\left(\sec \frac{x}{2}\right)^{2}$

$=2\left(\frac{1}{a^{2}\left(1+\left(\tan \frac{x}{2}\right)^{2}\right)+b^{2}\left(1+\left(\tan \frac{x}{2}\right)^{2}\right)+2 a b\left(1-\left(\tan \frac{x}{2}\right)^{2}\right)}\right) \sqrt{\left(a^{2}\right.}$

$\left.-b^{2}\right)\left(\sec \frac{x}{2}\right)^{2}$

Divide numerator and denominator by $\left(1+\left(\tan \frac{x}{2}\right)^{2}\right) ;$

We get:

$=2\left(\frac{1}{a^{2}+b^{2}+2 a b\left(\frac{1-\left(\tan \frac{x}{2}\right)^{2}}{1+\left(\tan \frac{x}{2}\right)^{2}}\right)}\right) \sqrt{\left(a^{2}-b^{2}\right)\left(\sec \frac{x}{2}\right)^{2} \frac{1}{1+\left(\tan \frac{x}{2}\right)^{2}}}$

$=2\left(\frac{1}{a^{2}+b^{2}+2 a b \cos x}\right) \sqrt{\left(a^{2}-b^{2}\right)\left(\sec \frac{x}{2}\right)^{2} \frac{1}{\left(\sec \frac{x}{2}\right)^{2}}}$

$=2\left(\frac{1}{a^{2}+b^{2}+2 a b \cos x}\right) \sqrt{\left(a^{2}-b^{2}\right)}$

$\frac{d^{2} y}{d x^{2}}=2 \sqrt{\left(a^{2}-b^{2}\right)\left(\frac{1}{a^{2}+b^{2}+2 a b \cos x}\right)^{2}\{-2 a b \sin x\}}$