If sin A, cos A are roots of the equation mx^2+nx+1=0, then find the r

1.	If sin A, cos A are roots of the equation mx^2+nx+1=0, then find the relation between m and n.
Answer» 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

If sin A, cos A are roots of the equation mx^2+nx+1=0, then find the relation between m and n.

Answer»

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