Question

Prove that:$(\sin A+\text{cosec}A{)}^2+(\cos A+\sec A{)}^2$ $=7+{\tan }^{2}A+{\cot }^{2}A$

Answer 1

LHS

$=(\sin A+\text{cosec}A{)}^2+(\cos A+\sec A{)}^2$

$={\sin }^{2}A+{\text{cosec}}^{2}A+2\sin A.\text{cosec}A+{\cos }^{2}A+{\sec }^{2}A+2\cos A.\sec A$

$={\sin }^{2}A+{\cos }^{2}A+2+2+{\text{cosec}}^{2}A+{\sec }^{2}A$

$=5+(1+{\cot }^{2}A)+(1+{\tan }^{2}A)$ ........ $[sin²\theta +cos²\theta =1]$, $[{\sec }^2\theta =1+{\tan }^2\theta ]$ and $[{cosec}^{2}x=1+{\cot }^{2}x]$

$=7+{\tan }^{2}A+{\cot }^{2}A$

= RHS

Hence, proved that

$(\sin A+\text{cosec}A{)}^2+(\cos A+\sec A{)}^2$ $=7+{\tan }^{2}A+{\cot }^{2}A$