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Prove the following identities(1 + tan α tan β)2 + (tan α – tan β)2 = sec2α sec2β |
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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. |
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