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Question
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
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Solution
डावी बाजू = `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)`
= `sintheta/(1/costheta + 1) + sintheta/(1/costheta - 1`
= `sintheta/((1 + costheta)/costheta) + sintheta/((1 - costheta)/(costheta))`
= `(sintheta costheta)/(1 + costheta) + (sintheta costheta)/(1 - costheta)`
= `sin theta costheta (1 /(1 + costheta) + 1/(1 - costheta))`
= `sintheta costheta [(1 - costheta + 1 + costheta)/((1 + costheta)(1 - costheta))]`
= `sintheta costheta (2/(1 - cos^2theta))` ......[∵ (a + b)(a – b) = a2 – b2]
= `sintheta costheta xx 2/(sin^2theta)` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
= `2 xx (costheta)/(sintheta)`
= 2cot θ
= उजवी बाजू
∴ `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
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cot2θ - tan2θ = cosec2θ - sec2θ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
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
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin 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.
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
