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प्रश्न
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
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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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संबंधित प्रश्न
secθ + tanθ = `cosθ/(1 - sinθ)`
sinθ × cosecθ = किती?
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
sec θ(1 - sin θ) (sec θ + tan θ) = 1
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
जर sec θ = `41/40`, तर sin θ, cot θ, cosec θ च्या किमती काढा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
