px^2 + qx + r = 0 has roots sin(theta) and cos(theta). Prove that p^2 - q^2 + 2pr = 0

The equation given is px^2 + qx + r = 0. Its roots are sinθ and cosθ.

For this equation,

Product of roots = r/p

=> sinθ.cosθ = r/p

Sum of roots = -q/p

=> sinθ + cosθ = -q/p           (because the roots are sinθ and cosθ)

Squaring both the sides, we'll get

=> (sinθ + cosθ)^2 = (-q/p)^2

=> sin^2 θ + cos^2 θ + 2sinθ.cosθ = q^2/p^2

=> 1 + 2sinθ.cosθ = q^2/p^2        (using sin^2 θ + cos^2 θ = 1)

Now, here, we will substitute the value of sinθ.cosθ we calculated in the beginning.

=> 1 + 2(r/p) = q^2/p^2

=> (p + 2r)/p = q^2/p^2

=> (p + 2r) = q^2/p

=> p^2 + 2pr = q^2

=> p^2 - q^2 + 2pr = 0

Hence Proved.