Question

If $\frac{\cos {12}^0+\sin {12}^0}{\cos {12}^0-\sin {12}^0}=\tan x^0$, what is the value of $x$?

Answer 1

$\frac{\cos {12}^0+\sin {12}^0}{\cos {12}^0-\sin {12}^0}=\frac{\cos {12}^0+\sin {12}^0}{\cos {12}^0}÷\frac{\cos {12}^0-\sin {12}^0}{\cos {12}^0}=\frac{1+\tan {12}^0}{1-\tan {12}^0}$
$=\tan ({45}^0+{12}^0)$
$=\tan {57}^0=\tan x^0$
x=57