Question

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

Answer 1

Solving the LHS of ${(\sin A+\csc A)}^2+{(\cos A+\sec A)}^2=7{\tan }^{2}A+{\cot }^{2}A$

${\left(\sin A+\csc A\right)}^2+{\left({\cos }^{2}A+{\sec }^{2}A\right)}^2={\sin }^{2}A+{\csc }^{2}A+2\sin A\csc A+{\cos }^{2}A+{\sec }^{2}A+2\cos A\sec A$$={\sin }^{2}A+{\cos }^{2}A+{\csc }^{2}A+{\sec }^{2}A+2\sin A\times \frac{1}{\sin A}+2\cos A\times \frac{1}{\cos A}$

$=1+\left(1+{\cot }^{2}A\right)+\left(1+{\tan }^{2}A\right)+2+2$

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

$=RHS$