Advertisements
Advertisements
प्रश्न
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
Advertisements
उत्तर
cosec A – sin A = p ......[दिलेले]
∴ `1/"sin A" - sin "A"` = p
∴ `(1 - sin^2"A")/"sin A"` = p
∴ `(cos^2"A")/"sin A"` = p ......`("i") [(because sin^2"A" + cos^2"A" = 1),(therefore 1 - sin^2"A" = cos^2"A")]`
sec A – cos A = q ......[दिलेले]
∴ `1/"cos A" - cos "A"` = q
∴ `(1 - cos^2"A")/"cos A"` = q
∴ `(sin^2"A")/"cos A"` = q .....(ii) `[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
डावी बाजू = `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)`
= `[((cos^2"A")/(sin "A"))^2 ((sin^2"A")/(cos"A"))]^(2/3) + [((cos^2"A")/(sin "A"))((sin^2"A")/(cos"A"))^2]^(2/3)` ......[(i) आणि (ii) वरून]
= `((cos^4"A")/(sin^2"A") xx (sin^2"A")/(cos"A"))^(2/3) + ((cos^2"A")/(sin"A") xx (sin^4"A")/(cos^2"A"))^(2/3)`
= `(cos^3"A")^(2/3) + (sin^3"A")^(2/3)`
= cos2A + sin2A
= 1
= उजवी बाजू
∴ `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
APPEARS IN
संबंधित प्रश्न
cos2θ(1 + tan2θ) = 1
खालील प्रश्नासाठी उत्तराचा योग्य पर्याय निवडा.
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
जर sec θ + tan θ = `sqrt(3)`, तर secθ – tanθ ची किंमत काढण्यासाठी खालील कृती पूर्ण करा.
कृती: `square` = 1 + tan2θ ......[त्रि. नित्य समीकरण]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
`(cos^2theta)/(sintheta) + sintheta` = cosec θ हे सिद्ध करा.
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= उजवी बाजू
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
सिद्ध करा:
cotθ + tanθ = cosecθ × secθ
उकल:
डावी बाजू = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
= उजवी बाजू
∴ cotθ + tanθ = cosecθ × secθ
