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
ANSWERED
asked 2021-09-02
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) \(\displaystyle{{\cos}^{{2}}{x}}{\tan{{x}}}={{\cos}^{{2}}{x}}{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}={\cos{{x}}}{\sin{{x}}}={2}{\sin{{2}}}{x}={\frac{{{1}}}{{{2}{\csc{{2}}}{x}}}}\)

Expert Answers (1)

2021-09-03
Solution:
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