Question

If $y=\sqrt{\frac{1-\cos 2x}{1+\cos 2x}},x\in \left(0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right)$, then $\frac{dy}{dx}$ is equal to

• A. $se{c}^{2}x,ifx\in \left(0,\frac{\pi }{2}\right)$
• B. $-se{c}^{2}x,ifx\in \left(\frac{\pi }{2},\pi \right)$
• C. $se{c}^{2}x,ifx\in \left(\frac{\pi }{2},\pi \right)$
• D. $-se{c}^{2}x,ifx\in \left(0,\frac{\pi }{2}\right)$

Answer 1

Answer: (A), (B)

$-se{c}^{2}x,ifx\in \left(\frac{\pi }{2},\pi \right)$

We have,
$y=\sqrt{\frac{1-\cos 2x}{1+\cos 2x}},x\in \left(0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right)\Rightarrow y=\sqrt{\frac{2{\sin }^{2}x}{2{\cos }^{2}x}}=\sqrt{{\tan }^{2}x}\Rightarrow y=\left(\tan x\right|,\mathrm{where}x\in \left(0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right)y=(\begin{array}{c}\tan x,\mathrm{if}x\in \left(0,\frac{\pi }{2}\right)\\ -\tan x,\mathrm{if}x\in \left(\frac{\pi }{2},\pi \right)\end{array}\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\frac{\mathrm{dy}}{\mathrm{dx}}=(\begin{array}{c}{\sec }^{2}x,\mathrm{if}x\in \left(0,\frac{\pi }{2}\right)\\ -{\sec }^{2}x,\mathrm{if}x\in \left(\frac{\pi }{2},\pi \right)\end{array}$