Question

Prove that $(1+{\tan }^{2}A)(1+\frac{1}{{\tan }^{2}A})=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$

Answer 1

Identity : ${\sec }^{2}A-{\tan }^{2}A=1$

$LHS=(1+{\tan }^{2}A)(1+\frac{1}{{\tan }^{2}A})$

$=\left({\sec }^{2}A\right)\frac{{\sec }^{2}A}{{\tan }^{2}A}$

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

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

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

$=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$

$=RHS$

Hence proved