Advertisements
Advertisements
प्रश्न
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
उत्तर
डावी बाजू = `sqrt((1 + cos "A")/(1 - cos"A"))`
= `sqrt((1 + cos "A")/(1 - cos "A") xx (1 + cos "A")/(1 + cos "A"))` ......[छेदाचे परिमेयकरण करून]
= `sqrt((1 + cos "A")^2/(1 - cos^2 "A"))`
= `sqrt((1 + cos "A")^2/(sin^2 "A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= `(1 + cos"A")/"sin A"`
= `1/"sin A" + "cos A"/"sin A"`
= cosec A + cot A
= उजवी बाजू
∴ `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
APPEARS IN
संबंधित प्रश्न
`sqrt((1 - sinθ)/(1 + sinθ))` = secθ - tanθ
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
sec2θ + cosec2θ = sec2θ × cosec2θ
जर 1 – cos2θ = `1/4`, तर θ = ?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
cotθ + tanθ = cosecθ × secθ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती:
डावी बाजू = cotθ + tanθ
= `costheta/sintheta + square/costheta`
= `(square + sin^2theta)/(sintheta xx costheta)`
= `1/(sintheta xx costheta)` ......`because square`
= `1/sintheta xx 1/costheta`
= `square xx sectheta`
डावी बाजू = उजवी बाजू
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
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.
