Question

proove

${\sec }^{2}A{\csc }^{2}A={\tan }^{2}A+{\cot }^{2}A+2$

Answer 1

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

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

$=\frac{{\sin }^{4}A+{\cos }^{4}A+2{\sin }^{2}A{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$

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

RHS$=\frac{1}{{\sin }^{2}A{\cos }^{2}A}$

$\therefore$LHS=RHS