If $sinx + siny = p, cosx + cosy = q$ then $\sec (x + y) =$

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$\frac{2 pq}{p^{2} + q^{2}}$

$\frac{p^{2} + q^{2}}{q^{2} - p^{2}}$

$\frac{p + q}{p^{2} + q^{2}}$

$\frac{2 pq}{\sqrt{p^{2} + q^{2}}}$

answer is B.

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Detailed Solution

It is given that, $sinx + siny = p and cosx + cosy = q$
So, on squaring and adding, we get $2 + 2 \cos (x - y) = p^{2} + q^{2} . ....... (1)$
On squaring and subtracting, we get $\begin{array}{l} \cos 2 x + cox 2 y + 2 \cos (x + y) = q^{2} - p^{2} \\ \Rightarrow 2 \cos (x + y) [\cos (x - y) + 1] = q^{2} - p^{2} . ....... (2) \end{array}$
From equations (1) and (2), we get $\sec (x + y) = \frac{p^{2} + q^{2}}{q^{2} - p^{2}}$