Question

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

Answer 1

Given:

$(1+\tan A\tan B{)}^2+(\tan A-\tan B{)}^2$

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

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

$={\sec }^{2}A+{\tan }^{2}B({\tan }^{2}A+1)$

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

$={\sec }^{2}A(1+{\tan }^{2}B)$

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