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Question
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
Solution
डावी बाजू = `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)`
= `1/("cosec" theta)(cottheta + tantheta)` .....`[(because tan(90 - theta) = cot theta),(cot(90 - theta) = tantheta)]`
= sin θ (cot θ + tan θ)
= `sintheta ((costheta)/(sintheta) + (sintheta)/(costheta))`
= `sintheta ((cos^2theta + sin^2theta)/(sintheta costheta))`
= `sintheta (1/(sintheta costheta))` ......[∵ sin2θ + cos2θ = 1]
= `1/costheta`
= sec θ
= उजवी बाजू
∴ `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
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sec2θ + cosec2θ = sec2θ × cosec2θ
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर cos θ = `24/25`, तर sin θ = ?
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B 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
