Question

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

Answer 1

${(\sin A+\csc A)}^2+{(\cos A+\sec A)}^2$

$={\sin }^{2}A+{\csc }^{2}A+2\sin A\csc A+{\cos }^{2}A+{\sec }^{2}A+2\cos A\sec A$ .......As$[a²+b²+2ab=(a+b\left)²\right]$

$={\sin }^{2}A+{\csc }^{2}A+2\sin A\times \frac{1}{\sin A}+{\cos }^{2}A+{\sec }^{2}A+2\cos A\frac{1}{\cos A}$ .

........... since $\sec A=\frac{1}{\cos A}$ and $\csc A=\frac{1}{\sin A}$

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

$=({\sin }^{2}A+{\cos }^{2}A)+{\csc }^{2}A+{\sec }^{2}A+4$

$=1+1+{\cot }^{2}A+1+{\tan }^{2}A+4$ ........... since ${\csc }^{2}A=1+{\cot }^{2}A$ and ${\sec }^{2}A=1+{\tan }^{2}A$

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

Hence proved.