Question

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

Answer 1

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

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

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

$={\tan }^2\left(A\right)[1+{\tan }^2(B\left)\right]+[1+{\tan }^2(B\left)\right]$

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

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