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प्रश्न
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
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उत्तर
डावी बाजू = `(sintheta + "cosec" theta)/sin theta`
= `sintheta/sintheta + ("cosec"theta)/sintheta`
= 1 + cosec θ × cosec θ ......`[∵ "cosec" theta = 1/sin theta]`
= 1 + cosec2θ
= 1 + 1 + cot2θ .......[∵ 1 + cot2θ = cosec2θ]
= 2 + cot2θ
= उजवी बाजू
∴ `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
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संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
sec4A(1 - sin4A) - 2tan2A = 1
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sec2θ + cosec2θ = sec2θ × cosec2θ
(sec θ + tan θ) . (sec θ – tan θ) = ?
`(cos^2theta)/(sintheta) + sintheta` = cosec θ हे सिद्ध करा.
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
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