We have,
$$
\begin{aligned}
& y=\frac{2}{\sqrt{a^2-b^2}} \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right] \\
& \Rightarrow \quad \frac{d y}{d x}=\left[\frac{2}{\sqrt{a^2-b^2}} \cdot \frac{1}{1+\left(\frac{a-b}{a+b}\right) \tan ^2 \frac{x}{2}}\right. \\
& \left.\cdot \sqrt{\frac{a-b}{a+b}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2}\right] \\
& =\frac{\sec ^2 x / 2}{a+b} \frac{a+b}{(a+b)+(a-b) \tan ^2 x / 2} \\
& =\frac{\sec ^2 x / 2}{a+b+a \tan ^2 \frac{x}{2}-b \tan ^2 x / 2} \\
& =\frac{\sec ^2 x / 2}{a\left(1+\tan ^2 \frac{x}{2}\right)+b\left(1-\tan ^2 \frac{x}{2}\right)} \\
& =\frac{\sec ^2 x / 2}{\left(1+\tan ^2 \frac{x}{2}\right)}\left[a+b\left(\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}\right)\right] \\
& \frac{d y}{d x}=\frac{1}{a+b \cos x} \\
& \therefore \quad \frac{d^2 y}{d x^2}=\frac{-1}{(a+b \cos x)^2} \cdot(-b \sin x) \\
& \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{b \sin x}{(a+b \cos x)^2} \\
& \left.\therefore \frac{d^2 y}{d x^2}\right|_{x=\frac{\pi}{2}}=\frac{b \sin \pi / 2}{\left(a+b \cos \frac{\pi}{2}\right)^2}=\frac{b}{a^2} \\
&
\end{aligned}
$$