# Prove the following trigonometric identities:

Taking RHS

tan6θ + 3tan2θsec2θ + 1

= (sec2θ - 1)3 + 3(sec2θ - 1)sec2θ + 1                  [As, tan2θ = sec2θ - 1]

= (sec6θ - 1 - 3sec4θ + 3sec2θ) + (3sec4θ - 3sec2θ) + 1              [(a + b)3 = a3 - b3 - 3a2b + 3ab2]

= sec6θ
= LHS
Hence Proved.

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