Question

Prove that ${\tan }^{2}A+{\cot }^{2}A={\sec }^{2}A.cose{c}^{2}A-2$

Answer 1

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

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

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

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

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

$=\frac{1}{co{s}^{2}A}.\frac{1}{si{n}^{2}A}-2$

$={\sec }^{2}A.cose{c}^{2}A-2$

$=RHS$

Hence proved