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 B $-\frac{1}{2}$

Let $t=\cos 2\theta$
$e^x=\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\Rightarrow e^x=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\Rightarrow e^x=\frac{1-\tan \theta }{1+\tan \theta }=\tan (\frac{\pi }{4}-\theta )\Rightarrow e^x\frac{dx}{d\theta }=-{\sec }^2(\frac{\pi }{4}-\theta )$

$\tan \frac{y}{2}=\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\Rightarrow \tan \frac{y}{2}=\tan \theta \Rightarrow \frac{1}{2}{\sec }^2\frac{y}{2}\frac{dy}{d\theta }={\sec }^2\theta$

When $t=\frac{1}{2}=\cos 2\theta \Rightarrow \theta =\frac{\pi }{6}$
$e^x=\tan (\frac{\pi }{4}-\frac{\pi }{6})\Rightarrow x=\ln \tan \frac{\pi }{12}$
$\tan \frac{y}{2}=\tan \frac{\pi }{6}\Rightarrow y=\frac{\pi }{3}$

So,
$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\Rightarrow \frac{dy}{dx}=\frac{2{\cos }^2\frac{y}{2}2{\sec }^2\theta }{-e^{-x}{\sec }^2(\frac{\pi }{4}-\theta )}$
Putting $t=\frac{1}{2},x=\ln \tan \frac{\pi }{12},y=\frac{\pi }{3}$
$\Rightarrow \frac{dy}{dx}=\frac{2{\cos }^2\frac{\pi }{6}2{\sec }^2\frac{\pi }{6}}{-e^{-\ln \tan \pi /12}{\sec }^2\left(\frac{\pi }{12}\right)}\Rightarrow \frac{dy}{dx}=\frac{2}{-\cot \frac{\pi }{12}{\sec }^2\left(\frac{\pi }{12}\right)}\Rightarrow \frac{dy}{dx}=-2\tan \frac{\pi }{12}{\cos }^2\frac{\pi }{12}\Rightarrow \frac{dy}{dx}=-\sin \frac{\pi }{6}=-\frac{1}{2}$