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प्रश्न
sec2θ + cosec2θ = sec2θ × cosec2θ
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उत्तर
डावी बाजू = sec2θ + cosec2θ
= `1/cos^2θ + 1/sin^2θ`
= `(sin^2θ + cos^2θ)/(cos^2θ . sin^2θ)`
= `1/(cos^2θ.sin^2θ)` .....[∵ sin2θ + cos2θ = 1]
= `1/cos^2θ xx 1/sin^2θ`
=`sec^2θ xx "cosec"^2θ`
= उजवी बाजू
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
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संबंधित प्रश्न
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
जर 1 – cos2θ = `1/4`, तर θ = ?
cot2θ × sec2θ = cot2θ + 1 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 sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
जर `1/sin^2θ - 1/cos^2θ-1/tan^2θ-1/cot^2θ-1/sec^2θ-1/("cosec"^2θ) = -3`, तर θ ची किमत काढा.
