Thank you for reporting, we will resolve it shortly
Q.
If $\tan \, \theta = \frac{\cos 25^{\circ} + \sin 25^{\circ} }{\cos 25^{\circ} - \sin 25^{\circ} }$ and $\theta$ is in the third quadrant, then $\theta = $
If $\tan \theta=\frac{\cos 25^{\circ}+\sin 25^{\circ}}{\cos 25^{\circ}-\sin 25^{\circ}}=\frac{1+\tan 25^{\circ}}{1-\tan 25^{\circ}}$
$=\tan \left(45^{\circ}+25^{\circ}\right)$
$\because \theta$ is in the third quadrant, so
$\tan \theta=\tan 70^{\circ}=\tan \left(180^{\circ}+70^{\circ}\right)=\tan 250^{\circ}$
$\Rightarrow \theta=250^{\circ} .$