Question -
Answer -
Let us consider LHS: (1 + tan α tan β) 2 +(tan α – tan β) 2
1+ tan2 α tan2 β + 2tan 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, 1 + tan2 θ = sec2 θ
So,
sec2 α sec2 β
= RHS
∴ LHS = RHS
Hence proved.