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Question
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
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Solution
डावी बाजू = `tanθ/(secθ - 1)`
= `tanθ/(secθ - 1) xx (secθ + 1)/(secθ + 1)` .......[छेदाचे परिमेयकरण करून]
= `(tanθ(secθ + 1))/(sec^2θ - 1)`
= `(tanθ(secθ + 1))/tan^2θ` .....`[(∵ 1 + tan^2θ = sec^2θ), (∴ sec^2θ - 1 = tan^2θ)]`
= `(secθ + 1)/tanθ`
∴ `tanθ/(secθ - 1) = (secθ + 1)/tanθ`
∴ समान गुणोत्तराच्या सिद्धांतानुसार,
`tanθ/(secθ - 1) = (secθ + 1)/tanθ`
= `(tanθ + (secθ + 1))/(secθ - 1 + (tanθ))`
= `(tanθ + secθ + 1)/(tanθ + secθ - 1)`
= उजवी बाजू
∴ `tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
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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
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
जर tan θ – sin2θ = cos2θ, तर sin2θ = `1/2` हे दाखवा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
