Question

If $\frac{1+\sin 2x}{1-\sin 2x}={\cot }^2(a+x)\forall x\in R\sim (n\pi +\frac{\pi }{4}),n\in N,$ then a can be

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. none of these

Answer 1

Answer: (D)

none of these

We have, $\frac{1+\sin 2x}{1-\sin 2x}$

$=\frac{1+\cos (2n\pi +\frac{\pi }{2}-2x)}{1-\cos (2n\pi +\frac{\pi }{2}-2x)}$

$=\frac{2{\cos }^2(n\pi +\frac{\pi }{4}-x)}{2{\sin }^2(n\pi +\frac{\pi }{4}-x)}$, where $n$ is any natural number.

$={\cot }^2(n\pi +\frac{\pi }{4}-x)$

$={\cot }^2(x-n\pi -\frac{\pi }{4})$

$={\cot }^2(a+x)$

hence can be any integer, putting $n=1$

we can see $a=\frac{5\pi }{4}$
Hence, option 'D' is correct.