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प्रश्न
`(cos^2theta)/(sintheta) + sintheta` = cosec θ हे सिद्ध करा.
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उत्तर
डावी बाजू = `(cos^2theta)/(sintheta) + sintheta`
= `(cos^2theta + sin^2theta)/sintheta`
= `1/sintheta` .......[∵ sin2θ + cos2θ = 1]
= cosec θ
= उजवी बाजू
∴ `(cos^2theta)/(sintheta) + sintheta` = cosec θ
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संबंधित प्रश्न
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
1 + tan2θ = किती?
sec2θ + cosec2θ = sec2θ × cosec2θ
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर 3 sin θ = 4 cos θ, तर sec 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
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
