Question

Proof trigonometric identities.\sec(x)+\tan(x)=\frac{\cos(x)}{1-\sin(x)}

Trigonometric equation and identitie
ANSWERED
asked 2021-09-12
Proof trigonometric identities.
\(\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}={\frac{{{\cos{{\left({x}\right)}}}}}{{{1}-{\sin{{\left({x}\right)}}}}}}\)

Expert Answers (1)

2021-09-13
The given equation is \(\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}={\frac{{{\cos{{\left({x}\right)}}}}}{{{1}-{\sin{{\left({x}\right)}}}}}}\)
Let choose \(\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}\) to prove \(\displaystyle{\frac{{{\cos{{x}}}}}{{{1}-{\sin{{x}}}}}}\) to prove the identity.
\(\displaystyle{\sec{{x}}}+{\tan{{x}}}={\frac{{{1}}}{{{\cos{{x}}}}}}+{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}\)
(\(\displaystyle\because{\sec{{x}}}={\frac{{{1}}}{{{\cos{{x}}}}}},{\tan{{x}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}\))
\(\displaystyle={\frac{{{1}+{\sin{{x}}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle={\frac{{{1}+{\sin{{x}}}}}{{{\cos{{x}}}}}}\cdot{\frac{{{\cos{{x}}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{{\cos}^{{2}}{x}}}}}\)
On further simplification,
\(\displaystyle{\sec{{x}}}+{\tan{{x}}}={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{\left({1}-{{\sin}^{{2}}{x}}\right)}}}}\)
\(\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{1}^{{2}}-{{\sin}^{{2}}{x}}}}}\)
\(\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{\left({1}+{\sin{{x}}}\right)}{\left({1}-{\sin{{x}}}\right)}}}}\)
\(\displaystyle={\frac{{{\cos{{x}}}}}{{{\left({1}-{\sin{{x}}}\right)}}}}\)
Hence, the given identity is proved.
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