Question

Prove that

${\tan }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B={\tan }^{2}A-{\tan }^{2}B$

Answer 1

LHS $={\tan }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B$

$=({\sec }^{2}A-1)(1+{\tan }^{2}B)-{\sec }^{2}A{\tan }^{2}B$ $[\because 1+{\tan }^2\theta ={\sec }^2\theta ]$

$={\sec }^{2}A+{\sec }^{2}A.{\tan }^{2}B-1-{\tan }^{2}B-{\sec }^{2}A.{\tan }^{2}B$

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

$={\sec }^{2}A-{\sec }^{2}B$

$=$RHS

Hence proved