Advertisements
Advertisements
प्रश्न
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
Advertisements
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
`(sin^2θ)/(cosθ) + cosθ = secθ`
cos2θ(1 + tan2θ) = 1
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
(sec θ + tan θ) (1 - sin θ) = cos θ
cot2θ - tan2θ = cosec2θ - sec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.
