Question

If $e^x=\frac{\sqrt{1+t}-\sqrt{1-t}}{\sqrt{1+t}+\sqrt{1-t}}$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$, then $\frac{dy}{dx}$ at $t=\frac{1}{2}$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. none of these

Answer 1

The correct option is A $-\frac{1}{2}$

Let $t=\cos 2\theta$
Then $e^x=\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}$
$=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{1-\tan \theta }{1+\tan \theta }=\tan (\frac{\pi }{4}-\theta )$
$\tan \frac{y}{2}=\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}=\tan \theta$
At $t=\frac{1}{2}$, $\cos 2\theta =\frac{1}{2}$ or
$\theta =\frac{\pi }{6}$
Then $x=\log \tan \frac{\pi }{12}$, $y=\frac{\pi }{3}$
Differentiating w.r.t. $\theta$, $e^x\frac{dx}{d\theta }=-{\sec }^2(\frac{\pi }{4}-\theta )$
and $\frac{1}{2}{\sec }^2\frac{y}{2}\frac{dy}{d\theta }={\sec }^2\theta$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{2{\sec }^2\theta {\cos }^2\frac{y}{2}}{-e^{-x}{\sec }^2\left(\frac{\pi }{4}-\theta \right)}$
At $t=\frac{1}{2}$, i.e.,
$\theta =\frac{\pi }{6}$, $\frac{dy}{dx}=\frac{2{\sec }^2\frac{\pi }{6}{\cos }^2\frac{\pi }{6}}{(-\left(e^{-\log \tan \frac{\pi }{12}}\right){\sec }^2\frac{\pi }{12})}$
$\therefore \frac{dy}{dx}=\frac{2}{-\cot \frac{\pi }{12}{\sec }^2\frac{\pi }{12}}=-2\tan \frac{\pi }{12}{\cos }^2\frac{\pi }{12}=-\sin \frac{\pi }{6}=-\frac{1}{2}$