Search any question & find its solution
Question:
Answered & Verified by Expert
If \(y=\sqrt{\frac{1+\tan x}{1-\tan x}}\), then \(\frac{d y}{d x}=\)
Options:
Solution:
1129 Upvotes
Verified Answer
The correct answer is:
\(\frac{1}{2}\left(\sqrt{\frac{1-\tan x}{1+\tan x}}\right) \sec ^2\left(\frac{\pi}{4}+x\right)\)
Given, \(y=\sqrt{\frac{1+\tan x}{1-\tan x}}\)
\(\begin{aligned}
\therefore \quad \frac{d y}{d x}= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{(1-\tan x) \sec ^2 x+(1+\tan x) \sec ^2 x}{(1-\tan x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \frac{2 \sec ^2 x}{(1-\tan x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \frac{2}{(\cos x-\sin x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \frac{1}{\left(\cos \left(x+\frac{\pi}{4}\right)\right)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \sec ^2\left(\frac{\pi}{4}+x\right)
\end{aligned}\)
Hence, option (a) is correct.
\(\begin{aligned}
\therefore \quad \frac{d y}{d x}= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}}\times \frac{(1-\tan x) \sec ^2 x+(1+\tan x) \sec ^2 x}{(1-\tan x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \frac{2 \sec ^2 x}{(1-\tan x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \frac{2}{(\cos x-\sin x)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \frac{1}{\left(\cos \left(x+\frac{\pi}{4}\right)\right)^2} \\
= & \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \sec ^2\left(\frac{\pi}{4}+x\right)
\end{aligned}\)
Hence, option (a) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.