Question

Prove the following identity :

${\displaystyle {\sec }^{2}A.\cos e{c}^{2}A={\tan }^{2}A+{\cot }^{2}A+2}$

Answer 1

LHS = ${\displaystyle {\sec }^{2}A.\cos e{c}^{2}A=\frac{1}{{\cos }^{2}A.{\sin }^{2}A}}$

RHS = ${\displaystyle {\tan }^{2}A+{\cot }^{2}A+2={\tan }^{2}A+{\cot }^{2}A+2{\tan }^{2}A.{\cot }^{2}A}$

 = ${\displaystyle {(\tan A+\cot A)}^2={(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A})}^2}$

 = ${\displaystyle {\left(\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A.\cos A}\right)}^2=\frac{1}{{\cos }^{2}A.{\sin }^{2}A}}$

= Hence , LHS = RHS