Find the angle between the tangents drawn from (3, 2) to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+4\mathrm{y}-2=0$

Equation of the given circle 
$\begin{array}{l}\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+4\mathrm{y}-2=0\\ \mathrm{r}=\sqrt{15}\end{array}$
$\text{ Let }\mathrm{P}=(3,2),\sqrt{{\mathrm{S}}_{11}}=\sqrt{9+4-18+8-2}=1$
If $\mathrm{\theta }$ is the angle between the tangents drawn from P to the circle S = 0 then
$\tan \frac{\mathrm{\theta }}{2}=\frac{\mathrm{r}}{\sqrt{{\mathrm{S}}_{11}}}=\sqrt{15}$
$\text{ Now }\cos \mathrm{\theta }=\frac{1-{\tan }^2\frac{\mathrm{\theta }}{2}}{1+{\tan }^2\frac{\mathrm{\theta }}{2}}=\frac{-7}{8}$
$\Rightarrow \mathrm{\theta }={\cos }^{-1}(-7/8)=\mathrm{\pi }-{\cos }^{-1}(7/8)$