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Question
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
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Solution
cosec A – sin A = p ...[Given]
∴ `1/(sin A) - sin A = p`
∴ `(1 - sin^2A)/(sin A) = p`
∴ `(cos^2A)/(sin A) = p` ...(i) `[(∵ sin^2A + cos^2A = 1),(∴ 1 - sin^2A = cos^2A)]`
sec A – cos A = q ...[Given]
∴ `1/(cos A) - cos A = q`
∴ `(1 - cos^2A)/(cos A) = q`
∴ `(sin^2A)/(cos A) = q` ...(ii) `[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
L.H.S. = `(p^2q)^(2/3) + (pq^2)^(2/3)`
= `[((cos^2A)/(sin A))^2 ((sin^2A)/(cos A))]^(2/3) + [((cos^2A)/(sin A))((sin^2A)/(cos A))^2]^(2/3)` ...[From (i) and (ii)]
= `((cos^4A)/(sin^2A) xx (sin^2A)/(cos A))^(2/3) + ((cos^2A)/(sin A) xx (sin^4A)/(cos^2A))^(2/3)`
= `(cos^3A)^(2/3) + (sin^3A)^(2/3)`
= cos2A + sin2A
= 1
= R.H.S.
∴ `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`
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