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 that,
$sin A + cos B = a$ .....(i)
and $sin B + cos A = b$ .....(ii)
On squaring both the equations and adding, we get
${sin}^{2} A + {cos}^{2} B + 2 sin A cos B + {sin}^{2} B + {cos}^{2} A + 2 sin B cos A = a^{2} + b^{2}$
$\Rightarrow 2 sin (A + B) + 2 = a^{2} + b^{2}$ .....As ${sin}^{2} θ + {cos}^{2} θ = 1$
$\Rightarrow sin A + B = \frac{a^{2} + b^{2} - 2}{2}$

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