If $sin A$ and $cos A$ are roots of the equation $p x^{2} + q x + m = 0$ , then the relation among $p, q$ and $m$ is :

q^{2} + m^{2} = {(p + m)}^{2}

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q^{2} - m^{2} = {(p + m)}^{2}

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q^{2} + m^{2} = {(p - m)}^{2}

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None of these

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Solution

The correct option is B $q^{2} + m^{2} = {(p + m)}^{2}$
Consider the given equation $p x^{2} + q x + m = 0$ whose roots are $sin A$ and

$cos A$

Then, we know that,

Sum of roots $= - \frac{c o f f i c e n t o f x}{c o f f i c e n t o f x^{2}}$

$sin A + cos A = - \frac{q}{p}$ ……… (1)

Now, multiple of roots $= \frac{c o n s tant t e r m}{c o f f i c e n t o f x^{2}}$

$sin A . cos A = \frac{m}{p}$ ………. (2)

On squaring both side equation (1) and we get,

${(sin A + cos A)}^{2} = {(- \frac{q}{p})}^{2}$

$\Rightarrow {sin}^{2} A + {cos}^{2} A + 2 sin A cos A = \frac{q^{2}}{p^{2}}$

$\Rightarrow 1 + 2 sin A cos = \frac{q^{2}}{p^{2}}$

$\Rightarrow 1 + 2 \frac{m}{p} = \frac{q^{2}}{p^{2}} b y e q u a t i o n (2)$

$\Rightarrow \frac{p + 2 m}{p} = \frac{q^{2}}{p^{2}}$

$\Rightarrow p (p + 2 m) = q^{2}$

$\Rightarrow p^{2} + 2 p m = q^{2}$

On adding both side $m^{2}$ and we get,

$p^{2} + 2 p m + m^{2} = q^{2} + m^{2}$

$\Rightarrow {(p + m)}^{2} = q^{2} + m^{2}$

Hence, this is the answer.

Option (A) is correct.

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