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Question
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
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Solution
डावी बाजू = cot2θ × sec2θ
= `(cos^2theta)/(sin^2theta) xx 1/(cos^2theta)`
= `1/(sin^2theta)`
= cosec2θ
= 1 + cot2θ ......[∵ 1 + cot2θ = cosec2θ]
= उजवी बाजू
∴ cot2θ × sec2θ = cot2θ + 1
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`(sin^2θ)/(cosθ) + cosθ = secθ`
`1/(secθ - tanθ)` = secθ + tanθ
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
sinθ × cosecθ = किती?
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
sec2θ + cosec2θ = sec2θ × cosec2θ
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
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
