Q. 314.0( 32 Votes )

# Prove the following trigonometric identities:

Answer :

Taking RHS

tan^{6}θ + 3tan^{2}θsec^{2}θ + 1

= (sec^{2}θ - 1)^{3 }+ 3(sec^{2}θ - 1)sec^{2}θ + 1 [As, tan^{2}θ = sec^{2}θ - 1]

= (sec^{6}θ - 1 - 3sec^{4}θ + 3sec^{2}θ) + (3sec^{4}θ - 3sec^{2}θ) + 1 [(a + b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}]

= sec^{6}θ

= LHS

Hence Proved.

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