If $\sin A$ and $\cos A$ are roots of the equation $p{x^2} + qx + m = 0$, then the relation among $p,q$ and $m$ is :
- ${q^2} + {m^2} = {\left( {p + m} \right)^2}$
- ${q^2} - {m^2} = {\left( {p + m} \right)^2}$
- ${q^2} + {m^2} = {\left( {p - m} \right)^2}$
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None of these
Roots are sin A and cos A. Sum = sin A + cos A = -q/p. Product = sin A cos A = m/p. Square the sum: sin^2 A + cos^2 A + 2 sin A cos A = q^2/p^2. 1 + 2(m/p) = q^2/p^2. p^2 + 2mp = q^2. Adding m^2 to both sides: p^2 + 2mp + m^2 = q^2 + m^2. (p+m)^2 = q^2 + m^2.
Using Vieta's formulas for the quadratic equation px^2 + qx + m = 0 with roots sin A and cos A, the sum of the roots gives (sin A + cos A) = -q/p and the product gives (sin A * cos A) = m/p. Squaring the sum equation yields (sin A + cos A)^2 = q^2/p^2, which expands to sin^2 A + cos^2 A + 2sin Acos A = q^2/p^2. Substituting the Pythagorean identity sin^2 A + cos^2 A = 1 and the product equation gives 1 + 2(m/p) = q^2/p^2. Multiplying by p^2 results in p^2 + 2pm = q^2, which can be rearranged as q^2 = p^2 + 2pm + m^2 - m^2 + 2pm - 2pm. This means q^2 + m^2 = p^2 + 2pm + m^2, which perfectly factors to q^2 + m^2 = (p + m)^2.