Question

If $y=\sqrt{\frac{1+\tan x}{1-\tan x}}$ then $\frac{dy}{dx}$ is equal to-

• A. $\frac{1}{2}\sqrt{\frac{1-\tan x}{1+\tan x}}{\sec }^2(\frac{\pi }{4}+x)$
• B. $\sqrt{\frac{1-\tan x}{1+\tan x}}{\sec }^2(\frac{\pi }{4}+x)$
• C. $\frac{1}{2}\sqrt{\frac{1-tanx}{1+\tan x}}\sec (\frac{\pi }{4}+x)$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}\sqrt{\frac{1-\tan x}{1+\tan x}}{\sec }^2(\frac{\pi }{4}+x)$

The given equation is:

$y=\sqrt{\frac{1+\tan x}{1-\tan x}}$

Differentiating w.r.t. to x once we get,

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{{\sec }^{2}x(1-\tan x)+{\sec }^{2}x(1+\tan x)}{(1-\tan x{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{2{\sec }^{2}x}{{\sec }^{2}x-2\tan x}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{1}{\frac{1}{2}-\sin x\cos x}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{1}{\frac{1}{2}({\sin }^{2}x+{\cos }^{2}x)-2.\frac{1}{\sqrt{2}}\sin x.\frac{1}{\sqrt{2}}\cos x}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{1}{{(-\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}})}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{1}{{\cos }^2(\frac{\pi }{4}+x)}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\times \sqrt{\frac{1-\tan x}{1+\tan x}}\times {\sec }^2(\frac{\pi }{4}+x)$ .....Answer