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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
जर tanθ = 2, तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा
sec6x - tan6x = 1 + 3sec2x × tan2x
(sec θ + tan θ) . (sec θ – tan θ) = ?
`(sin^2theta)/(cos theta) + cos theta` = sec θ 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]`
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
