Question

If $\cot (x+y)=\frac{1}{\sqrt{3}}$ and $\cot (x-y)=\sqrt{3}$, then find the smallest positive value of $x$ and $y$.

Answer 1

$\tan (x+y)=\sqrt{3}$ and $\tan (x-y)=\frac{1}{\sqrt{3}}$
$\therefore \tan (x+y)=\tan {60}^0$ or $\tan {240}^0$
and $\tan (x-y)=\tan {30}^0$ or $\tan {210}^0$
$\therefore x+y={60}^0$ and $x-y={30}^0$
so$x={45}^0$ $y={15}^0$