Question

If $\frac{1+\sin 2x}{1-\sin 2x}={\cot }^2(a+x)$, for all $x\in R-(n\pi +\frac{\pi }{4})n\in N$. Then $a$ is equal to (Given that $a\in (0,\pi )$)

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{\pi }{8}$

Answer 1

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

According to question...................

$\begin{array}{c}\frac{1+sin2x}{1-sin2x}=co{t}^2(a+x)\\ L.H.S:\frac{1+sin2x}{1-sin2x}[\begin{array}{c}fromformula:{\sin }^{2}x+{\cos }^{2}x=1\\ \sin 2x=2\sin x\cos x\end{array}\\ \Rightarrow \frac{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}\\ ifbothdivideby\cos x,\\ \Rightarrow \frac{{\left(\frac{\cos x+\sin x}{\cos x}\right)}^2}{{\left(\frac{\cos x-\sin x}{\cos x}\right)}^2}={\left[\frac{1+\tan x}{1-\tan x}\right]}^2\\ \Rightarrow {\left[\frac{1}{\frac{1-\tan x}{1+\tan x}}\right]}^2={\left[\frac{1}{\frac{-(\tan x-1)}{1+\tan x}}\right]}^2\\ \Rightarrow {\left[\frac{1}{\frac{(\tan x-1)}{1+\tan x}}\right]}^2={\left[\frac{1}{\frac{(\tan x+\tan (\frac{3\pi }{4})}{1-(-1)\tan x}}\right]}^2[\tan \left(\frac{3\pi }{4}\right)=-1]\\ \Rightarrow {\left[\frac{1}{\frac{(\tan x+\tan (\frac{3\pi }{4})}{1-\tan \left(\frac{3\pi }{4}\right)\tan x}}\right]}^2[\tan (A+B)=\frac{\tan A+\tan B}{1-tanA\tan B}]\\ now,\\ L.H.S:{\left[\frac{1}{\tan \left(\frac{3\pi }{4}+x\right)}\right]}^2\\ \Rightarrow {\cot }^2\left(\frac{3\pi }{4}+x\right)\\ wehavegiven,R.H.S:{\cot }^2(a+x)\\ Now,comparetoL.H.S\&R.H.S\\ weget:a=\frac{3\pi }{4}\\ sothatthecorrectoptionisC.\end{array}$