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If $\sin x+\sin y=p, \cos x+\cos y=q$, then $\sec (x+y)=$
Options:
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Verified Answer
The correct answer is:
$\frac{p^2+q^2}{q^2-p^2}$
It is given that, $\sin x+\sin y=p$, and
$\cos x+\cos y=q$
So,
on squaring and adding, we get
$2+2 \cos (x-y)=p^2+q^2$
On squaring and subtracting, we get
$\cos 2 x+\cos 2 y+2 \cos (x+y)=q^2-p^2$
$\Rightarrow 2 \cos (x+y)[\cos (x-y)+1]=q^2-p^2$
From Eqs. (i) and (ii), we get
$\sec (x+y)=\frac{p^2+q^2}{q^2-p^2}$
Hence, option (b) is correct.
$\cos x+\cos y=q$
So,
on squaring and adding, we get
$2+2 \cos (x-y)=p^2+q^2$
On squaring and subtracting, we get
$\cos 2 x+\cos 2 y+2 \cos (x+y)=q^2-p^2$
$\Rightarrow 2 \cos (x+y)[\cos (x-y)+1]=q^2-p^2$
From Eqs. (i) and (ii), we get
$\sec (x+y)=\frac{p^2+q^2}{q^2-p^2}$
Hence, option (b) is correct.
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