Question

Prove the identity. \frac{\sin(2x)}{\sin(x)}-\frac{\cos(2x)}{\cos(x)}=\sec(x)

Trigonometric equation and identitie
ANSWERED
asked 2021-09-11
Prove the identity
\(\displaystyle{\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{\sin{{\left({x}\right)}}}}}}-{\frac{{{\cos{{\left({2}{x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}}={\sec{{\left({x}\right)}}}\)

Expert Answers (1)

2021-09-12
Solution:
\(\displaystyle{\frac{{{\sin{{\left({2}{x}\right)}}}}}{{{\sin{{\left({x}\right)}}}}}}-{\frac{{{\cos{{\left({2}{x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}}={\sec{{\left({x}\right)}}}\)
\(\displaystyle{\sin{{2}}}{x}={2}{\sin{{x}}}{\cos{{x}}}\)
\(\displaystyle{{\cos}^{{2}}{2}}{x}={{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}\)
\(\displaystyle={\frac{{{2}{\sin{{x}}}{\cos{{x}}}}}{{{\sin{{x}}}}}}-{\frac{{{{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle={2}{\cos{{x}}}-{\frac{{{\left({{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}\right)}}}{{{\cos{{x}}}}}}\)
Taking LCH
\(\displaystyle={\frac{{{2}{{\cos}^{{2}}{x}}-{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle={\frac{{{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)
\(\displaystyle{\frac{{{1}}}{{{\cos{{x}}}}}}={\sec{{x}}}\)
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