Prove the following identities (1 + tan α tan β) 2 + (tan α – tan β) 2 = sec 2 α sec 2 β

Question

Philoid · Accepted Answer

LHS = (1 + tan α tan β)² + (tan α – tan β)²

= 1+ tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

= 1 + tan² α tan² β + tan² α + tan² β

= tan² α (tan² β + 1) + 1 (1 + tan² β)

= (1 + tan² β) (1 + tan² α)

We know that 1 + tan² θ = sec² θ

= sec² α sec² β

= RHS

Hence proved.