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प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
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उत्तर
डावी बाजू = `sqrt((1 - sinθ)/(1 + sinθ))`
= `sqrt((1 - sinθ)/(1 + sinθ) xx (1 - sinθ)/(1 - sinθ))` .....[छेदाचे परिमेयकरण करून]
= `sqrt((1 - sinθ)^2/(1 - sin^2θ)`
= `sqrt((1 - sinθ)^2/cos^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1), (∴ 1 - sin^2θ = cos^2θ)]`
= `(1 - sinθ)/(cosθ) = 1/cosθ - sinθ/cosθ`
= sec θ - tan θ
= उजवी बाजू
∴ `sqrt((1 - sinθ)/(1 + sinθ))` = sec θ - tan θ
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संबंधित प्रश्न
sec4θ - cos4θ = 1 - 2cos2θ
sec4A(1 - sin4A) - 2tan2A = 1
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
जर 3 sin θ = 4 cos θ, तर sec θ = ?
tan2θ – sin2θ = tan2θ × sin2θ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
`sec"A"/(tan "A" + cot "A")` = sin A हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
