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Question
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
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Solution
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
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RELATED QUESTIONS
1 + tan2θ = किती?
tan4θ + tan2θ = sec4θ - sec2θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
जर tan θ = `7/24`, तर cos θ ची किंमत काढण्यासाठी खालील कृती पूर्ण करा.
कृती: sec2θ = 1 + `square` ......[त्रि. नित्य समीकरण]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
cot θ + tan θ = cosec θ × sec θ, हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती:
डावी बाजू = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= उजवी बाजू
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "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.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
जर tan θ – sin2θ = cos2θ, तर sin2θ = `1/2` हे दाखवा.
