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प्रश्न
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
उत्तर
डावी बाजू = sin2A . tan A + cos2A . cot A + 2 sin A . cos A
= `sin^2"A"* (sin "A")/(cos "A") + cos^2"A"* (cos"A")/(sin"A") + 2sin"A" *cos"A"`
= `(sin^3"A")/"cosA" + (cos^3"A")/"sinA" + 2sin"A"*cos"A"`
= `(sin^4"A" + cos^4"A" + 2sin^2"A"cos^2"A")/(sin"A"cos"A")`
= `(sin^2"A" + cos^2"A")^2/(sin"A"cos"A")` .....[∵ a2 + b2 + 2ab = (a + b)2]
= `1^2/(sin"A"cos"A")` ......[∵ sin2A + cos2A = 1]
= `1/(sin"A"cos"A")`
= `(sin^2"A"+ cos^2"A")/(sin"A"cos"A")` ......[∵ 1 = sin2A + cos2A]
= `(sin^2"A")/(sin"A"cos"A") + (cos^2"A")/(sin"A"cos"A")`
= `"sinA"/"cosA" + "cosA"/"sinA"`
= tan A + cot A
= उजवी बाजू
∴ sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
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