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Prove the identity: (1 + tan α tan β)2 + (tan α – tan β)2 = sec2 α sec2 β

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Let us consider the 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)

As we know, 1 + tan2 θ = sec2 θ

Therefore,

sec2 α sec2 β

= RHS

∴ LHS = RHS

Thus proved.

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