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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
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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
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संबंधित प्रश्न
cos2θ(1 + tan2θ) = 1
(sec θ + tan θ) (1 - sin θ) = cos θ
`(tan^3θ - 1)/(tanθ - 1)` = sec2θ + tanθ
(sec θ + tan θ) . (sec θ – tan θ) = ?
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
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
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
