Question

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

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

Answer 1

Answer: (C)

$\tan \left(1^o\right)+\tan \left({89}^o\right)=\tan \left(1^o\right)+\tan ({90}^o-1^o)=\tan \left(1^o\right)+\cot \left(1^o\right)$

$=\frac{\sin \left(1^o\right)}{\cos \left(1^o\right)}+\frac{\cos \left(1^o\right)}{\sin \left(1^o\right)}$

$=\frac{{\sin }^2\left(1^o\right)+{\cos }^2\left(1^o\right)}{\sin \left(1^o\right)\cos \left(1^o\right)}=\frac{2}{\sin 2^o}$