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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
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
sec4A(1 - sin4A) - 2tan2A = 1
(sec θ + tan θ) . (sec θ – tan θ) = ?
जर sec θ + tan θ = `sqrt(3)`, तर secθ – tanθ ची किंमत काढण्यासाठी खालील कृती पूर्ण करा.
कृती: `square` = 1 + tan2θ ......[त्रि. नित्य समीकरण]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
sec2θ − cos2θ = tan2θ + sin2θ हे सिद्ध करा.
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]`
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
cotθ + tanθ = cosecθ × secθ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती:
डावी बाजू = cotθ + tanθ
= `costheta/sintheta + square/costheta`
= `(square + sin^2theta)/(sintheta xx costheta)`
= `1/(sintheta xx costheta)` ......`because square`
= `1/sintheta xx 1/costheta`
= `square xx sectheta`
डावी बाजू = उजवी बाजू
