Question

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

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

Answer 1

Answer: (B)

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