Advertisements
Advertisements
प्रश्न
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
Advertisements
उत्तर
डावी बाजू = `(1 + sec "A")/"sec A"`
= `1/"sec A" + "sec A"/"sec A"`
= cos A + 1
= `(1 + cos "A") xx (1 - cos"A")/(1 - cos"A")`
= `(1 - cos^2"A")/(1 - cos"A")`
= `(sin^2"A")/(1 - cos"A")` .......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= उजवी बाजू
∴ `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
APPEARS IN
संबंधित प्रश्न
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
(sec θ + tan θ) (1 - sin θ) = cos θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
sec2θ – tan2θ = ?
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
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
