Question

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

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

Answer 1

The correct option is B $\frac{2}{\sin \left(2^{\circ }\right)}$

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

$=\frac{\sin 1^o}{\cos 1^o}+\frac{\cos 1^o}{\sin 1^o}=\frac{{\sin }^2{1}^o+{\cos }^2{1}^o}{\sin 1^o\cos 1^o}=\frac{1}{\sin 1^o\cos 1^o}$

$=\frac{2}{2\sin 1^o\cos 1^o}=\frac{2}{\sin 2^o}$

Formula used: ${\sin }^2\theta +{\cos }^2\theta =1$ and $\sin 2\theta =2\sin \theta \cos \theta$

Also $\cot ({90}^o-\theta )=\cot \theta$