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प्रश्न
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
उत्तर
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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संबंधित प्रश्न
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sin2θ + sin2(90 – θ) = ?
`costheta/(1 + sintheta) = (1 - sintheta)/(costheta)` हे सिद्ध करा.
`(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ हे सिद्ध करा.
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 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
