Question

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

Answer 1

Now,

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

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

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

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

$={\sec }^{2}A.{\sec }^{2}B$. [ $1+{\tan }^{2}A={\sec }^{2}A$ and $1+{\tan }^{2}B={\sec }^{2}B$]