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Question
sec θ(1 - sin θ) (sec θ + tan θ) = 1
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Solution
डावी बाजू = sec θ(1 - sin θ) (sec θ + tan θ)
= `1/cos θ(1 - sin θ)(1/cos θ + sin θ/cos θ)`
= `(1 - sin θ)/cos θ((1 + sin θ)/cos θ)`
= `(1 - sin^2θ)/(cos^2θ)`
= `(cos^2θ)/(cos^2θ)` ..........`[(∵ sin^2θ + cos^2θ = 1), (∴ 1 - sin^2θ = cos^2θ)]`
= 1
= उजवी बाजू
∴ sec θ(1 - sin θ) (sec θ + tan θ) = 1
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जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
(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
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
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
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
