Question

$\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$ is equal to

• A. 1
• B. -1
• C. 0
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Given, $ \frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$
$=\frac{1-\tan 13^{\circ}}{1+\tan 13^{\circ}}+\tan 148^{\circ}$
$=\frac{\tan 45^{\circ}-\tan 13^{\circ}}{1+\tan 45^{\circ} \tan 13^{\circ}}+\tan 148^{\circ}$
$=\tan \left(45^{\circ}-13^{\circ}\right)+\tan \left(180^{\circ}-32^{\circ}\right)$
$=\tan 32^{\circ}-\tan 32^{\circ}=0$