Q. 14 D

Prove the following identities :

tan4θ + tan2θ = sec4θ – sec2θ

Answer :

Taking LHS = tan4 θ + tan2 θ

= (tan2 θ)2 + tan2 θ


= ( sec2 θ – 1)2 + (sec2 θ – 1) [ 1+ tan2 θ = sec2 θ ]


= sec4 θ + 1 – 2 sec2 θ + sec2 θ – 1 [ (a b)2 = (a2 + b2 – 2ab)]


= sec4 θ – sec2 θ


=RHS


Hence Proved


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