Question

If $\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$ then the value of ${\tan }^2\left(\frac{\theta -\varphi }{2}\right)$ is

Answer 1

Applying the trigonometric identities in $\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$.

$2\sin \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=a$ (1)

And

$2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=b$ (2)

Square and add equation (1) and (2),

$4{\sin }^2\frac{\theta +\varphi }{2}{\cos }^2\frac{\theta -\varphi }{2}+4{\cos }^2\frac{\theta +\varphi }{2}{\cos }^2\frac{\theta -\varphi }{2}=a^2+b^2$

$4{\cos }^2\frac{\theta -\varphi }{2}\left({\sin }^2\frac{\theta +\varphi }{2}+{\cos }^2\frac{\theta +\varphi }{2}\right)=a^2+b^2$

$4{\cos }^2\frac{\theta -\varphi }{2}=a^2+b^2$

${\sec }^2\frac{\theta -\varphi }{2}=\frac{4}{a^2+b^2}$

$1+{\tan }^2\frac{\theta -\varphi }{2}=\frac{4}{a^2+b^2}$

${\tan }^2\frac{\theta -\varphi }{2}=\frac{4}{a^2+b^2}-1$

${\tan }^2\frac{\theta -\varphi }{2}=\frac{4-a^2-b^2}{a^2+b^2}$