Advertisements
Advertisements
Question
`costheta/(1 + sintheta) = (1 - sintheta)/(costheta)` हे सिद्ध करा.
Solution
डावी बाजू = `costheta/(1 + sintheta)`
= `costheta/(1 + sintheta) xx (1 - sintheta)/(1 - sintheta)` ......[छेदाचे परिमेयकरण करून]
= `(costheta(1 - sintheta))/(1 - sin^2theta)`
= `(costheta(1 - sintheta))/(cos^2theta)` ......`[(because sin^2theta +cos^2theta = 1),(therefore 1 -sin^2theta = cos^2theta)]`
= `(1 - sintheta)/costheta`
= उजवी बाजू
∴ `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
APPEARS IN
RELATED QUESTIONS
जर tanθ + `1/tanθ` = 2 तर दाखवा की `tan^2θ + 1/tan^2θ` = 2
sec2θ + cosec2θ = sec2θ × cosec2θ
cot2θ - tan2θ = cosec2θ - sec2θ
`tanθ/(secθ + 1) = (secθ - 1)/tanθ`
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
(sec θ + tan θ) . (sec θ – tan θ) = ?
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
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
`sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.