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प्रश्न
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
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उत्तर
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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संबंधित प्रश्न
cot θ + tan θ = cosec θ sec θ
secθ + tanθ = `cosθ/(1 - sinθ)`
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
sec θ(1 - sin θ) (sec θ + tan θ) = 1
sec6x - tan6x = 1 + 3sec2x × tan2x
जर 1 – cos2θ = `1/4`, तर θ = ?
sec2θ + cosec2θ = sec2θ × cosec2θ हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` 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
सिद्ध करा:
cotθ + tanθ = cosecθ × secθ
उकल:
डावी बाजू = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
= उजवी बाजू
∴ cotθ + tanθ = cosecθ × secθ
