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प्रश्न
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
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उत्तर
डावी बाजू = sin4A – cos4A
= (sin2A)2 – (cos2A)2
= (sin2A + cos2A)(sin2A – cos2A) .....[∵ a2 – b2 = (a + b)(a – b)]
= (1)(sin2A – cos2A) ......[∵ sin2A + cos2A = 1]
= sin2A – cos2A
= (1 – cos2A) – cos2A ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"" = sin^2"A")]`
= 1 – 2cos2A
= उजवी बाजू
∴ sin4A – cos4A = 1 – 2cos2A
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संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
cot θ + tan θ = cosec θ sec θ
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
जर 1 – cos2θ = `1/4`, तर θ = ?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
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
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
