If $sin x + cos x = a$ then $| sin x - cos x | =$

\sqrt{1 - a^{2}}

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\sqrt{a^{2} - 1}

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\sqrt{2 - a^{2}}

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\sqrt{a^{2} - 2}

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Solution

The correct option is C $\sqrt{2 - a^{2}}$
$\begin{matrix} sin x + cos x = a s q u a r i n g b o t h s i d e s {(sin x + cos x)}^{2} = a^{2} {sin}^{2} x + {cos}^{2} x + 2 sin x cos x = a^{2} 1 + 2 sin x cos x = a^{2} 2 sin x cos x = a^{2} - 1 N o w, {(sin x - cos x)}^{2} = {sin}^{2} x + {cos}^{2} x - 2 sin x cos x {(sin x - cos x)}^{2} = 1 - 2 sin x cos x {(sin x - cos x)}^{2} = 1 - a^{2} + 1 {(sin x - cos x)}^{2} = 2 - a^{2} sin x - cos x = \sqrt{2 - a^{2}} \end{matrix}$
Hence, the option $C$ is the correct answer.

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