Question

If $\sin \alpha +\sin \beta =a$ and $\cos \alpha +\cos \beta =b$, prove that $\tan \frac{\alpha -\beta }{2}=\pm \sqrt{\frac{x-a^2-b^2}{a^2+b^2}}$. Find $x$

Answer 1

Given, $\sin \alpha +\sin \beta =a$
and $\cos \alpha +\cos \beta =b$
Now ${(\cos \alpha +\cos \beta )}^2+(\sin \alpha +\sin \beta )=b^2+a^2$
or ${\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta =b^2+a^2$
or $({\cos }^2\alpha +{\sin }^2\alpha )+({\cos }^2\beta +{\sin }^2\beta )+2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=a^2+b^2$
or $2+2\cos (\alpha -\beta )=a^2+b^2$
or $\cos (\alpha -\beta )=\frac{a^2+b^2-2}{2}$
Now, $\tan \frac{\alpha -\beta }{2}=\pm \sqrt{\frac{1-\cos (\alpha -\beta )}{1-\cos (\alpha -\beta )}}$
$=\pm \sqrt{\frac{1-\frac{a^2+b^2-2}{2}}{1+\frac{a^2+b^2-2}{2}}}=\pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$
Ans: 4