Question

2) Prove that (1+tan A tan B)2 + (tan A
+ (tan A - tan B)2=
sec2A sec2 B​

Answer 1

LHS = (1 + tan a tan b)²+ (tan a – tan b)²

= 1 + tan²a tan²b + 2 tan a tan b + tan²a + tan²b – 2 tan a tan b

= 1 + tan²a tan²b + tan²a + tan²b

= 1. (1 + tan²b) + tan²a (tan²b + 1)

= (1 + tan²b) + (1 + tan²a)

= sec²b× sec²a

= sec2a sec2b

= RHS