Q. 134.0( 19 Votes )

Prove the following identities

(1 + tan α tan β)2 + (tan α – tan β)2 = sec2 α sec2 β

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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