The angle between the pair of tangents from the point $$\left( {1,\,\frac{1}{2}} \right)$$  to the circle $${x^2} + {y^2} + 4x + 2y - 4 = 0$$      is :

Solution :
Equation of pair tangents to the circle $${x^2} + {y^2} + 4x + 2y - 4 = 0$$      from the point $$\left( {1,\,\frac{1}{2}} \right)$$  is
$$\eqalign{ & S{S_1} = {T^2}\left( {{x^2} + {y^2} + 4x + 2y - 4} \right)\left( {1 + \frac{1}{4} + 4 + 1 - 4} \right) \cr & = {\left( {x + \frac{1}{2}y + 2\left( {x + 1} \right)y + \frac{1}{2} - 4} \right)^2} \cr & \Rightarrow \frac{9}{4}\left( {{x^2} + {y^2} + 4x + 2y - 4} \right) = {\left( {3x + \frac{3}{2}y - \frac{3}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4}{x^2} + \frac{9}{4}{y^2} + 9x + \frac{9}{2}y - 9 = 9{x^2} + \frac{9}{4}{y^2} + \frac{9}{4} + 9xy - \frac{9}{2}y - 9x \cr & \Rightarrow \frac{{27}}{4}{x^2} - 18x - 9y + 9xy + \frac{{45}}{4} = 0 \cr & \Rightarrow \frac{3}{4}{x^2} + xy - 2x - y + \frac{5}{4} = 0 \cr & \Rightarrow 3{x^2} + 4xy - 8x - 4y + 5 = 0 \cr} $$
It $$\theta $$ is the angle of these lines, then
$$\eqalign{ & \tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}} \cr & \Rightarrow \tan \,\theta = \frac{{2\sqrt 4 }}{3} \cr & \Rightarrow \tan \,\theta = \frac{4}{3} \cr & \Rightarrow \sin \,\theta = \frac{4}{5} \cr & \Rightarrow \theta = {\sin ^{ - 1}}\left[ {\frac{4}{5}} \right] \cr} $$