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प्रश्न
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
उत्तर
डावी बाजू = `tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2`
= `tanA/(sec^2A)^2 + cotA/(cosec^2A)^2` .........`[(∵ 1 + tan^2θ = sec^2θ), (∴ 1 + cot^2θ = cosec^2θ)]`
= `tanA/sec^4A + cotA/(cosec^4A)`
= `tanA xx 1/sec^4A + cotA xx 1/(cosec^4A)`
= `sinA/cosA xx cos^4A + cosA/sinA xx sin^4A`
= sin A cos3A + cos A sin3A
= sin A cos A(cos2A + sin2A)
= sin A cos A (1) ........[∵ sin2θ + cos2θ = 1]
= sin A cos A
= उजवी बाजू
∴ `tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
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संबंधित प्रश्न
sec4A(1 - sin4A) - 2tan2A = 1
`tanθ/(secθ - 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
sinθ × cosecθ = किती?
जर sinθ = `11/61`, तर नित्यसमानतेचा उपयोग करून cosθ ची किंमत काढा.
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
जर 1 – cos2θ = `1/4`, तर θ = ?
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ 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
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
