Question

If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)

Answer 1

LHS = (b - c)(r - p) = `(b - asin^2θ - bcos^2θ)(p sin^2θ + qcos^2θ - p)`

= `[b(1 - cos^2θ) - asin^2θ][p(sin^2θ - 1) + q cos^2θ]`

⇒ LHS = `[(b - a)sin^2θ][(q - p)cos^2θ] = (b - a)(q - p)sin^2θcos^2θ`

RHS = `(c - a)(q- r) = (asin^2θ + bcos^2θ - a)(q - p sin^2θ - qcos^2θ)`

= `[(b - a)cos^2θ][(q - p)sin^2θ] = (b - a)(q - p)sin^2θ.cos^2θ`

Thus , (b - c)(r - p) = (c - a)(q - r)