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प्रश्न
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
उत्तर
sin6A + cos6A = (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" sin^2"A")]`
= 1 – 3 cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ......[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
sin4A + cos4A = (sin2A)2 + (cos2A)2
= (1 – cos2A)2 + (cos2A)2
= 1 – 2 cos2A + (cos2A)2 + (cos2A)2 ......[∵ (a – b)2 = a2 – 2ab + b2]
= 1 – 2 cos2A + 2 cos4A
= 1 – 2 cos2A(1 – cos2A)
= 1 – 2 cos2A sin2A
डावी बाजू = 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1
= 2(1 – 3 cos2A sin2A) – 3(1 – 2 cos2A sin2A) + 1
= 2 – 6 cos2A sin2A – 3 + 6 cos2A sin2A + 1
= 0
= उजवी बाजू
∴ 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
APPEARS IN
संबंधित प्रश्न
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
`1/(secθ - tanθ)` = secθ + tanθ
sec4A(1 - sin4A) - 2tan2A = 1
sec2θ + cosec2θ = sec2θ × cosec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sin2θ + sin2(90 – θ) = ?
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
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
