Question

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

Answer 1

LHS

Expand LHS

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

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

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

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

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

$\Rightarrow {\sec }^{2}A{\sec }^{2}B$ Hence proved