Question

# Prove the identity , \frac{1}{2\csc 2x}=\cos^2 x \tan x . Choose the sequence of steps below that verifies the identity

Trigonometric equation and identitie
Prove the identity
$$\displaystyle{\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}={{\cos}^{{2}}{x}}{\tan{{x}}}$$
Choose the sequence of steps below that verifies the identity
A) $$\displaystyle{{\cos}^{{2}}{x}}{\tan{{x}}}={{\cos}^{{2}}{x}}{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}={\cos{{x}}}{\sin{{x}}}={\frac{{{\sin{{2}}}{x}}}{{{2}}}}={\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}$$
B) $$\displaystyle{{\cos}^{{2}}{x}}{\tan{{x}}}={{\cos}^{{2}}{x}}{\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}={\cos{{x}}}{\sin{{x}}}={\frac{{{\sin{{2}}}{x}}}{{{2}}}}={\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}$$
C) $$\displaystyle{{\cos}^{{2}}{x}}{\tan{{x}}}={{\cos}^{{2}}{x}}{\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}={\cos{{x}}}{\sin{{x}}}={2}{\sin{{2}}}{x}={\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}$$
D) $$\displaystyle{{\cos}^{{2}}{x}}{\tan{{x}}}={{\cos}^{{2}}{x}}{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}={\cos{{x}}}{\sin{{x}}}={2}{\sin{{2}}}{x}={\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}$$