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प्रश्न
sec4θ - cos4θ = 1 - 2cos2θ
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उत्तर
डावी बाजू = sec4θ - cos4θ
= (sec2θ)2 – (cos2θ)2
= (sec2θ + cos2θ) (sec2θ – cos2θ) ....[∵ a2 – b2 = (a + b)(a – b)]
`= ((1 + cos^4 theta)(1- cos^4 theta))/(cos^4 theta)`
= `(1 + cos^4 theta)(1- cos^4 theta) (1 + cos^2 theta)/cos^4 theta`
Thus the solution would be not coming equal to RHS.
The correct question would be sin4 θ in place of sec4θ.
On solving this question we get,
= (sin2θ)2 – (cos2θ)2
= (sin2θ + cos2θ) (sin2θ – cos2θ) ....[∵ a2 – b2 = (a + b)(a – b)]
= (1) (sin2θ – cos2θ) ....[∵ sin2θ + cos2θ = 1]
= sin2θ – cos2θ
= (1 - cos2θ) - cos2θ ....[∵ sin2θ = 1 - cos2θ]
= 1 - 2cos2θ = उजवी बाजू
∴ sin4 θ - cos4θ = 1 - 2 cos2θ
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संबंधित प्रश्न
`1/(secθ - tanθ)` = secθ + tanθ
sec4A(1 - sin4A) - 2tan2A = 1
sinθ × cosecθ = किती?
sec2θ + cosec2θ = sec2θ × cosec2θ
cot2θ - tan2θ = cosec2θ - sec2θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
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
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
