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प्रश्न
Prove that sin4A – cos4A = 1 – 2 cos2A.
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उत्तर
L.H.S. = sin4A – cos4A
= (sin2A)2 – (cos2A)2
= (sin2A + cos2A)(sin2A – cos2A) ...[∵ a2 – b2 = (a + b)(a – b)]
= (1)(sin2A – cos2A) ...[∵ sin2A + cos2A = 1]
= sin2A – cos2A
= (1 – cos2A) – cos2A ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= 1 – 2 cos2A
= R.H.S.
∴ sin4A – cos4A = 1 – 2 cos2A
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
