Question

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

• A. $\frac{2 p q}{p^{2}+q^{2}}$
• B. $\frac{p^{2}+q^{2}}{q^{2}-p^{2}}$
• C. $\frac{2 p q}{\sqrt{p^{2}+q^{2}}}$
• 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
$\cos \,2 x+\cos \,2 y+2 \,\cos (x+y)=q^{2}-p^{2}$
$\Rightarrow 2 \cos (x+y)[\cos (x-y)+1]=q^{2}-p^{2}\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$\sec (x+y)=\frac{p^{2}+q^{2}}{q^{2}-p^{2}}$