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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
जर sin θ = `11/61`, तर नित्यसमानतेचा उपयोग करून cos θ ची किंमत काढा.
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
sec6x - tan6x = 1 + 3sec2x × tan2x
cosec θ.`sqrt(1 - cos^2theta) = 1` हे सिद्ध करा.
जर 1 – cos2θ = `1/4`, तर θ = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
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
sin6A + cos6A = 1 – 3sin2A . cos2A 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.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
