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प्रश्न
sec4A(1 - sin4A) - 2tan2A = 1
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उत्तर
डावी बाजू = sec4A(1 - sin4A) - 2tan2A
= sec4A[12 – (sin2A)2] – 2tan2A
= sec4A .(1 – sin2A) (1 + sin2A) – 2tan2A
= sec4A cos2A (1 + sin2A) – 2tan2A ...`[(∵ sin^2θ + cos^2θ = 1), (∴ 1 - sin^2θ = cos^2θ)]`
= `1/cos^4A . cos^2A(1 + sin^2A) - 2tan^2A`
= `1/cos^2A (1 + sin^2A) - 2tan^2A`
= `1/cos^2A + sin^2A/cos^2A - 2tan^2A`
= sec2A + tan2A – 2tan2A
= sec2A – tan2A
= 1 ................[∵ sec2θ – tan2θ = 1]
= उजवी बाजू
∴ sec4A(1 - sin4A) - 2tan2A = 1
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संबंधित प्रश्न
जर sin θ = `11/61`, तर नित्यसमानतेचा उपयोग करून cos θ ची किंमत काढा.
sec2θ + cosec2θ = sec2θ × cosec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
(sec θ + tan θ) . (sec θ – tan θ) = ?
tan2θ – sin2θ = tan2θ × sin2θ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
