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प्रश्न
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
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उत्तर
डावी बाजू = `(sin^2theta)/(cos theta) + cos theta`
= `(sin^2theta + cos^2theta)/costheta`
= `1/costheta` ......[∵ sin2θ + cos2θ = 1]
= sec θ
= उजवी बाजू
∴ `(sin^2theta)/(cos theta) + cos theta` = sec θ
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संबंधित प्रश्न
`(sin^2θ)/(cosθ) + cosθ = secθ`
cos2θ(1 + tan2θ) = 1
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
sec6x - tan6x = 1 + 3sec2x × tan2x
`(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)`
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
जर 1 – cos2θ = `1/4`, तर θ = ?
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
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.
sin2θ + cos2θ ची किंमत काढा.
उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
