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`.

Answer 1

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`