Question

Solve
${(1+\tan a\tan b)}^2+(\tan a-\tan b)={\sec }^{2}a{\sec }^{2}b$

Answer 1

$(1+\tan a+\tan b{)}^2+(\tan a-\tan b{)}^2$

$=1+2\tan a\tan b+{\tan }^{2}a{\tan }^{2}b+{\tan }^{2}a+{\tan }^{2}b-2\tan a\tan b$

$=1+{\tan }^{2}a+{\tan }^{2}a{\tan }^{2}b+{\tan }^{2}b$

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

$={\sec }^{2}a+{\tan }^{2}b{\sec }^{2}a$

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

$=se{c}^{2}a{\sec }^{2}b$.