Question

$\cot {660}^o+\tan \left({-1050}^o\right)=?$

Answer 1

Consider the following Equation.

$\cot {660}^0+\tan (-{1050}^0)=\cot (90\times 8-60)-\tan (90\times 12-30)$

$=-\cot {60}^0+\tan {30}^0$

$=-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}$

$=0$

Hence, this is the correct answer.