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प्रश्न
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
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उत्तर
L.H.S. = sin2A . tan A + cos2A . cot A + 2 sin A . cos A
= `sin^2A * (sin A)/(cos A) + cos^2A * (cos A)/(sin A) + 2 sin A * cos A`
= `(sin^3A)/(cos A) + (cos^3A)/(sin A) + 2 sin A * cos A`
= `(sin^4A + cos^4A + 2 sin^2A cos^2A)/(sinA cosA)`
= `(sin^2A + cos^2A)^2/(sinA cosA)` ...[∵ a2 + b2 + 2ab = (a + b)2]
= `1^2/(sinA cosA)` ...[∵ sin2A + cos2A = 1]
= `1/(sinA cosA)`
= `(sin^2A + cos^2A)/(sinA cosA)` ...[∵ 1 = sin2A + cos2A]
= `(sin^2A)/(sinA cosA) + (cos^2A)/(sinA cosA)`
= `(sin A)/(cos A) + (cos A)/(sin A)`
= tan A + cot A
= R.H.S.
∴ sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
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Solution :
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∴ cotθ + tanθ = cosecθ × secθ
