Question

If $\tan \theta =\frac{\cos 25^{\circ }+\sin 25^{\circ }}{\cos 25^{\circ }-\sin 25^{\circ }}$ and $\theta$ is in the third quadrant, then $\theta =$

• A. $200^{\circ }$
• B. $205^{\circ }$
• C. $225^{\circ }$
• D. $250^{\circ }$

Answer 1

Answer: (D)

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
$\tan \theta =\tan 70^{\circ }=\tan \left(180^{\circ }+70^{\circ }\right)=\tan 250^{\circ }$
$\Rightarrow \theta =250^{\circ }.$