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प्रश्न
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
उत्तर
डावी बाजू = `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1)`
= `(cot"A" + "cosec A" - ("cosec"^2"A" - cot^2"A"))/(cot"A" - "cosec A" + 1)` .....`[(because 1 + cot^2"A" = "cosec"^2"A"),(therefore "cosec"^2"A" - cot^2"A" = 1)]`
= `(cot"A" + "cosec A" - ("cosec A" + cot"A")("cosec A" - cot"A"))/(cot"A" - "cosec A" + 1)` .....[∵ a2 – b2 = (a + b) (a – b)]
= `((cot"A" + "cosec A")(1 - "cosec A" + cot "A"))/(cot"A" - "cosec A" + 1)`
= cot A + cosec A
= `"cos A"/"sin A" + 1/"sin A"`
= `(cos "A" + 1)/"sin A"`
= उजवी बाजू
∴ `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
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संबंधित प्रश्न
cot θ + tan θ = cosec θ sec θ
`1/(secθ - tanθ)` = secθ + tanθ
secθ + tanθ = `cosθ/(1 - sinθ)`
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
sec θ(1 - sin θ) (sec θ + tan θ) = 1
cot θ + tan θ = cosec θ × sec θ, हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती:
डावी बाजू = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= उजवी बाजू
जर sec θ = `41/40`, तर sin θ, cot θ, cosec θ च्या किमती काढा.
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
