Question

The value of tan $\left(1^{\circ}\right)+$ tan $\left(89^{\circ}\right)$ is equal to

• A. $\frac{1}{\sin (1^{\circ})}$
• B. $\frac{2}{\sin (2^{\circ})}$
• C. $\frac{2}{\sin (1^{\circ})}$
• D. $\frac{1}{\sin (2^{\circ})}$
• E. $\frac{2} {\sin (2^{\circ})}$

Answer 1

Answer: (B)

$ \tan 1^{\circ}+\tan 89^{\circ}=\tan 1^{\circ}+\tan \left(90^{\circ}-1^{\circ}\right) $
$= \tan 1^{\circ}+\cot 1^{\circ}=\frac{\sin 1^{\circ}}{\cos 1^{\circ}}+\frac{\cos 1^{\circ}}{\sin 1^{\circ}}$
$= \frac{\sin ^{2} 1^{\circ}+\cos ^{2} 1^{\circ}}{\sin 1^{\circ} \cos 1^{\circ}} $
$= \frac{2}{2 \sin 1^{\circ} \cos 1^{\circ}}=\frac{2}{\sin 2^{\circ}} $