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