Question

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

• A. $\frac{b\cos x}{{(a-b\sin x)}^2}$
• B. $\frac{b\cos x}{{(a+b\sin x)}^2}$
• C. $\frac{b\sin x}{{(a-b\cos x)}^2}$
• D. $\frac{b\sin x}{{(a+b\cos x)}^2}$

Answer 1

The correct option is D $\frac{b\sin x}{{(a+b\cos x)}^2}$

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

Then $AB=\frac{2}{\sqrt{(a-b)(a+b)}}\sqrt{\frac{a-b}{a+b}}\Rightarrow AB=\frac{2}{a+b}$

$\frac{dy}{dx}=\frac{AB\frac{1}{2}{\sec }^2\frac{x}{2}}{1+B^2{\tan }^2\frac{x}{2}}=\frac{1}{a+b}\times \frac{(a+b)}{(a+b){\cos }^2\frac{x}{2}+(a-b){\sin }^2\frac{x}{2}}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{a+b\cos x}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{b\sin x}{{(a+b\cos x)}^2}$