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प्रश्न
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
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उत्तर
डावी बाजू = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(bb(cos^2theta + sin^2theta))/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `bb([sin^2theta + cos^2theta = 1])`
= `1/sinθ xx 1/bbcostheta`
= cosecθ × secθ
= उजवी बाजू
∴ cotθ + tanθ = cosecθ × secθ
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संबंधित प्रश्न
cot θ + tan θ = cosec θ sec θ
`1/(secθ - tanθ)` = secθ + tanθ
secθ + tanθ = `cosθ/(1 - sinθ)`
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
