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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
APPEARS IN
संबंधित प्रश्न
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
sinθ × cosecθ = किती?
sec2θ + cosec2θ = sec2θ × cosec2θ
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= उजवी बाजू
tan2θ – sin2θ = tan2θ × sin2θ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
