Question

Q: Prove that: $\dfrac{{1 + \sin 2x + \cos 2x}}{{1 + \sin 2x - \cos 2x}} = \cot x$

Answer 1

Hint: Start with left hand side, use formulas $1 + \cos 2x = 2{\cos ^2}x$, $1 - \cos 2x = 2{\sin ^2}x$ and $\sin 2x = 2\sin x\cos x$ and then simplify it further to bring it in the form of right hand side.

Complete step by step answer:
From the question,
$
   \Rightarrow LHS = \dfrac{{1 + \sin 2x + \cos 2x}}{{1 + \sin 2x - \cos 2x}} \\
   \Rightarrow LHS = \dfrac{{(1 + \cos 2x) + \sin 2x}}{{(1 - \cos 2x) + \sin 2x}} \\
$
We know that $1 + \cos 2x = 2{\cos ^2}x$, $1 - \cos 2x = 2{\sin ^2}x$ and $\sin 2x = 2\sin x\cos x$. Using these formulas for above expression, we’ll get:
$
   \Rightarrow LHS = \dfrac{{2{{\cos }^2}x + 2\sin x\cos x}}{{2{{\sin }^2}x + 2\sin x\cos x}} \\
   \Rightarrow LHS = \dfrac{{2\cos x(\cos x + \sin x)}}{{2\sin x(\sin x + \cos x)}} \\
   \Rightarrow LHS = \dfrac{{\cos x}}{{\sin x}} \\
   \Rightarrow LHS = \cot x = RHS \\
$
This is the required proof.

Note: The formula for $\cos 2x$ can be used in three different forms:
$
   \Rightarrow \cos 2x = 2{\cos ^2}x - 1 \\
   \Rightarrow \cos 2x = 1 - 2{\sin ^2}x \\
   \Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x \\
$
We can use any of them as per the requirement of the question.