Question

If $y=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right]$, then $\left.\frac{d^{2} y}{d x^{2}}\right|_{x=\frac{\pi}{2}}=$

• A. $\frac{b}{2 a^{2}}$
• B. $\frac{b}{a^{2}}$
• C. $\frac{2 b}{a}$
• D. $\frac{b^{2}}{2 a}$

Answer 1

Answer: (B)

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