Question

If $\sin \alpha +\sin \beta =l,\cos \alpha +\cos \beta =m$ and $\tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)=n(\ne 1)$, then

• A. $\cos (\alpha -\beta )=\frac{l^2+m^2-2}{2}$
• B. $\cos (\alpha +\beta )=\frac{m^2-l^2}{m^2+l^2}$
• C. $\frac{1+n}{1-n}=\frac{l^2+m^2}{2n}$
• D. $\alpha +\beta =\frac{\pi }{2}$ if $l=m$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\cos (\alpha -\beta )=\frac{l^2+m^2-2}{2}$
B $\cos (\alpha +\beta )=\frac{m^2-l^2}{m^2+l^2}$
D $\alpha +\beta =\frac{\pi }{2}$ if $l=m$

$\sin \alpha +\sin \beta =l\Rightarrow l^2={\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta$ ...(1)
$\cos \alpha +\cos \beta =m\Rightarrow m^2={\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta$ ...(2)
Adding (1) and (2)
$2\cos (\alpha -\beta )=l^2+m^2-2$ ...(3)
Subtracting (2) from (1)
$\cos 2\alpha +\cos 2\beta +2\cos (\alpha +\beta )=m^2-l^2\Rightarrow 2\cos (\alpha +\beta )\cos (\alpha -\beta )+2\cos (\alpha +\beta )=m^2-l^2\Rightarrow \cos (\alpha +\beta )\left(2\cos \right(\alpha -\beta )+2)=m^2-1^2$
$\Rightarrow \cos (\alpha +\beta )=\frac{m^2-1^2}{m^2+l^2}$ ...(4)
Now
$\frac{1+n}{1-n}=\frac{1+\tan \alpha \tan \beta }{1-\tan \alpha \tan \beta }=\frac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }$
$=\frac{\cos (\alpha -\beta )}{\cos (\alpha +\beta )}=\frac{(l^2+m^2-2).(m^2+l^2)}{m^2-1^2}$
If $l=m$
$\cos (\alpha +\beta )=0\Rightarrow \alpha +\beta =\frac{\pi }{2}$