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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
tan4θ + tan2θ = sec4θ - sec2θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
cos2θ . (1 + tan2θ) = 1 हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `cos^2theta xx square` .........`[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= उजवी बाजू
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
sec2θ − cos2θ = tan2θ + sin2θ हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
जर `1/sin^2θ - 1/cos^2θ-1/tan^2θ-1/cot^2θ-1/sec^2θ-1/("cosec"^2θ) = -3`, तर θ ची किमत काढा.
