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Question
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
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Solution
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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RELATED QUESTIONS
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cot2θ - tan2θ = cosec2θ - sec2θ
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
जर sec θ = `41/40`, तर sin θ, cot θ, cosec θ च्या किमती काढा.
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
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`
