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Question
sec4A(1 - sin4A) - 2tan2A = 1
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Solution
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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`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
sec2θ + cosec2θ = sec2θ × cosec2θ
tan4θ + tan2θ = sec4θ - sec2θ
(sec θ + tan θ) . (sec θ – tan θ) = ?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= उजवी बाजू
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
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
