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प्रश्न
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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उत्तर
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = sec6 θ
= (sec2 θ)3 = (1 + tan2 θ)3
= 1 + (tan2 θ)3 + 3 (1) (tan2 θ) (1 + tan2 θ) ......[(a + b)3 = a3 + b3 + 3 ab (a + b)]
= 1 + tan6 θ + 3 tan2 θ (1 + tan2 θ)
= 1 + tan6 θ + 3 tan2 θ (sec2 θ)
= 1 + tan6 θ + 3 tan2 θ sec2 θ
= tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = R.H.S
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संबंधित प्रश्न
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Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
