Question

If $\sin x+\sin y=p,\cos x+\cos y=q$, then $\sec (x+y)=$

• A. $\frac{2pq}{p^2+q^2}$
• B. $\frac{p^2+q^2}{q^2-p^2}$
• C. $\frac{2pq}{\sqrt{p^2+q^2}}$
• D. $\frac{p+q}{p^2+q^2}$

Answer 1

Answer: (B)

It is given that, $\sin x+\sin y=p$, and
$\cos x+\cos y=q$
So, on squaring and adding, we get
$2+2\cos (x-y)=p^2+q^2...(i)$
On squaring and subtracting, we get
$\cos 2x+\cos 2y+2\cos (x+y)=q^2-p^2$
$\Rightarrow 2\cos (x+y)[\cos (x-y)+1]=q^2-p^2...(ii)$
From Eqs. (i) and (ii), we get
$\sec (x+y)=\frac{p^2+q^2}{q^2-p^2}$