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प्रश्न
If cosec A – sin A = p and sec A – cos A = q, then prove that `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
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उत्तर
cosec A – sin A = p ......[Given]
∴ `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 ......[Given]
∴ `1/"cos A" - cos "A"` = q
∴ `(1 - cos^2"A")/"cos A"` = q
∴ `(sin^2"A")/"cos A"` = q .....`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
L.H.S = `("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)` ......[From (i) and (ii)]
= `((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
= R.H.S
∴ `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
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