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Prove the following trigonometric identities:sec6 θ = tan6 θ + 3tan2 θsec2 θ + 1 |
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Answer» 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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