Question

If $\frac{1+\sin 2x}{1-\sin 2x}={\cot }^2(a+x)$$\forall x\in R-(n\pi +\frac{\pi }{4}),$ $n\in N$ then the possible value of $a$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is C $\frac{3\pi }{4}$

$\frac{1+\sin 2x}{1-\sin 2x}=\frac{(\sin x+\cos x{)}^2}{(\sin x-\cos x{)}^2}$
$={\left(\frac{1+\tan x}{1-\tan x}\right)}^2={\left(\tan (\frac{\pi }{4}+x)\right)}^2={\cot }^2(\frac{\pi }{2}+\frac{\pi }{4}+x)(\because \cot (\frac{\pi }{2}+x)=-\tan x)={\cot }^2(\frac{3\pi }{4}+x)\Rightarrow {\cot }^2(a+x)={\cot }^2(\frac{3\pi }{4}+x)\therefore a=\frac{3\pi }{4}$