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प्रश्न
`tanA/(1 + tan^2A)^2 + cotA/(1 + cot^2A)^2` = sin A cos A
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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
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
sec4θ - cos4θ = 1 - 2cos2θ
secθ + tanθ = `cosθ/(1 - sinθ)`
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot 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.
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
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.
