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प्रश्न
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
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उत्तर
डावी बाजू = `sqrt((1 + cos "A")/(1 - cos"A"))`
= `sqrt((1 + cos "A")/(1 - cos "A") xx (1 + cos "A")/(1 + cos "A"))` ......[छेदाचे परिमेयकरण करून]
= `sqrt((1 + cos "A")^2/(1 - cos^2 "A"))`
= `sqrt((1 + cos "A")^2/(sin^2 "A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= `(1 + cos"A")/"sin A"`
= `1/"sin A" + "cos A"/"sin A"`
= cosec A + cot A
= उजवी बाजू
∴ `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
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संबंधित प्रश्न
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
(sec θ + tan θ) (1 - sin θ) = cos θ
sec2θ + cosec2θ = sec2θ × cosec2θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
जर 1 – cos2θ = `1/4`, तर θ = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
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
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
