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प्रश्न
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
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
cos2θ(1 + tan2θ) = 1
1 + tan2θ = किती?
sec2θ + cosec2θ = sec2θ × cosec2θ
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर cos θ = `24/25`, तर sin θ = ?
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
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
सिद्ध करा:
cotθ + tanθ = cosecθ × secθ
उकल:
डावी बाजू = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
= उजवी बाजू
∴ cotθ + tanθ = cosecθ × secθ
sin2θ + cos2θ ची किंमत काढा.
उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
