If $sin A + cos B = a$ and $sin B + cos A = b$ , then $sin (A + B)$ is equal to

\frac{a^{2} + b^{2}}{2}

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\frac{a^{2} - b^{2} + 2}{2}

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\frac{a^{2} + b^{2} - 2}{2}

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None of these

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Solution

The correct option is B $\frac{a^{2} + b^{2} - 2}{2}$
Given, $sin A + cos B = a$
and $sin B + cos A = b$
On squaring and adding, we get
${sin}^{2} A + {cos}^{2} B + 2 sin A cos B + {sin}^{2} B + {cos}^{2} A + 2 sin B cos A$
$= a^{2} + b^{2}$
$\Rightarrow ({sin}^{2} A + {cos}^{2} A) + ({cos}^{2} B + {sin}^{2} B) + 2 (sin A cos B + cos A sin B)$
$= a^{2} + b^{2}$
$\Rightarrow 1 + 1 + 2 sin (A + B) = a^{2} + b^{2}$
$\Rightarrow sin (A + B) = \frac{a^{2} + b^{2} - 2}{2}$ .

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