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प्रश्न
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
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उत्तर
डावी बाजू = (sec θ - cos θ)(cot θ + tan θ)
= `(1/cos θ - cos θ) (cos θ/sin θ + sin θ/cos θ)`
= `((1 - cos^2θ)/cos θ)((cos^2θ + sin^2θ)/(sinθcosθ))`
= `sin^2θ/cosθ xx 1/(sinθcosθ)` ....`[(∵ sin^2θ + cos^2θ = 1),(∴ sin^2θ = 1 - cos^2θ)]`
= `sinθ/cosθ . 1/cosθ`
= tan θ . sec θ
= उजवी बाजू
∴ (sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
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संबंधित प्रश्न
sec4θ - cos4θ = 1 - 2cos2θ
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
sec4A(1 - sin4A) - 2tan2A = 1
sinθ × cosecθ = किती?
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
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
