Question

Verify that the equation is an identity.\frac{\cos\theta+1}{\tan^2\theta}=\frac{\cos\theta}{\sec\theta-1}

Trigonometric equation and identitie
ANSWERED
asked 2021-08-31
Verify that the equation is an identity.
\(\displaystyle{\frac{{{\cos{\theta}}+{1}}}{{{{\tan}^{{2}}\theta}}}}={\frac{{{\cos{\theta}}}}{{{\sec{\theta}}-{1}}}}\)

Expert Answers (1)

2021-09-01
Solution:
\(\displaystyle{\frac{{{\cos{\theta}}+{1}}}{{{{\tan}^{{2}}\theta}}}}={\frac{{{\cos{\theta}}}}{{{\sec{\theta}}-{1}}}}\)
\(\displaystyle{R}{H}{S}={\frac{{{\cos{\theta}}}}{{{\sec{\theta}}-{1}}}}\times{\frac{{{\sec{\theta}}+{1}}}{{{\sec{\theta}}-{1}}}}\)
\(\displaystyle={\frac{{{\cos{\theta}}{\left({\sec{\theta}}+{1}\right)}}}{{{{\sec}^{{2}}\theta}-{1}}}}\)
\(\displaystyle={\frac{{{\cos{\theta}}{\sec{\theta}}+{\cos{\theta}}}}{{{{\tan}^{{2}}\theta}}}}\)
\(\displaystyle={\frac{{{\cos{\cdot}}{\frac{{{1}}}{{{\cos{\theta}}}}}+{\cos{\theta}}}}{{{{\tan}^{{2}}\theta}}}}={\frac{{{1}+{\cos{\theta}}}}{{{{\tan}^{{2}}\theta}}}}\)
\(\displaystyle={\frac{{{\cos{\theta}}+{1}}}{{{{\tan}^{{2}}\theta}}}}={L}{H}{S}\)
RHS=LHS
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