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Question
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
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Solution
उजवी बाजू = `(sec^2"A")/("cosec"^2"A")`
= `(1 + tan^2"A")/(1 + cot^2"A")` .....`[(because 1 + tan^2"A" = sec^2"A"),(1 + cot^2"A" = "cosec"^2A")]`
= `(1 + (sin^2"A")/(cos^2"A"))/(1 + (cos^2"A")/(sin^2"A"))`
= `((cos^2"A" + sin^2"A")/(cos^2"A"))/((sin^2"A" + cos^2"A")/(sin^2"A"))`
= `(1/(cos^2"A"))/(1/(sin^2"A"))` .......[∵ sin2A + cos2A = 1]
= `(sin^2"A")/(cos^2"A")`
= tan2A
= tan A . tan A
= `"tan A"/"cot A"`
= डावी बाजू
∴ `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
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`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
खालीलपैकी चुकीचे सूत्र कोणते?
जर sec θ = `41/40`, तर sin θ, cot θ, cosec θ च्या किमती काढा.
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
जर tan θ – sin2θ = cos2θ, तर sin2θ = `1/2` हे दाखवा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
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
