Advertisements
Advertisements
प्रश्न
`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.
Advertisements
उत्तर
उजवी बाजू = `(sec^2"A")/("cosec"^2"A")`
= `(1 + tan^2"A")/(1 + cot^2"A")` .....`[(because 1 + tan^2"A" = sec^2"A"),(1 + cot^2"A" = "cosec"^2A")]`
= `(1 + (sin^2"A")/(cos^2"A"))/(1 + (cos^2"A")/(sin^2"A"))`
= `((cos^2"A" + sin^2"A")/(cos^2"A"))/((sin^2"A" + cos^2"A")/(sin^2"A"))`
= `(1/(cos^2"A"))/(1/(sin^2"A"))` .......[∵ sin2A + cos2A = 1]
= `(sin^2"A")/(cos^2"A")`
= tan2A
= tan A . tan A
= `"tan A"/"cot A"`
= डावी बाजू
∴ `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
APPEARS IN
संबंधित प्रश्न
sec θ(1 - sin θ) (sec θ + tan θ) = 1
(sec θ + tan θ) (1 - sin θ) = cos θ
sec2θ + cosec2θ = sec2θ × cosec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
cos2θ . (1 + tan2θ) = 1 हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `cos^2theta xx square` .........`[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= उजवी बाजू
जर tan θ = `7/24`, तर cos θ ची किंमत काढण्यासाठी खालील कृती पूर्ण करा.
कृती: sec2θ = 1 + `square` ......[त्रि. नित्य समीकरण]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
