Easy trigonometry problem

This problem can be solved using half-angle tangent substitution as follows:

$\begin{aligned} \sec 50^\circ + \tan 10^\circ & = \frac 1{\cos (30^\circ + 20^\circ)} + \tan 10^\circ \\ & = \frac 2{\sqrt 3\cos 20^\circ - \sin 20^\circ} + \tan 10^\circ & \small \blue{\text{Let }t=\tan 10^\circ} \\ & = \frac {2(1+t^2)}{\sqrt 3(1-t^2) - 2t} + t \\ & = \frac {2+\sqrt 3t-\sqrt 3t^3}{\sqrt 3-2t - \sqrt 3 t^2} \\ & = \frac {2 - 2\sqrt 3 t + \blue{\sqrt 3(3t-t^3)}}{\sqrt 3-2t - \sqrt 3 t^2} & \small \blue{\text{As }\tan 30^\circ = \frac {3t-t^3}{1-3t^2} = \frac 1{\sqrt 3}} \\ & = \frac {2 - 2\sqrt 3 t + \blue{1-3t^2}}{\sqrt 3-2t - \sqrt 3 t^2} \\ & = \frac {\sqrt 3(\cancel{\sqrt 3-2t - \sqrt 3 t^2})}{\cancel{\sqrt 3-2t - \sqrt 3 t^2}} \\ & = \boxed{\sqrt 3} \end{aligned}$

$\sec 50^{\circ}+\tan 10^{\circ}$ $=\frac{ \cos 10^{\circ}+\sin 10^{\circ} \sin 40^{\circ}}{ \sin 40^{\circ} \cos 10^{\circ}}$ $=\frac{ \cos (60^{\circ}-50^{\circ})-\frac{\cos 50^{\circ}}{2} +\frac{\cos 30^{\circ}}{2}}{ \frac{\sin 50^{\circ}}{2}+\frac{\sin 30^{\circ}}{2}}$ $=\frac{ \frac{\cos 50^{\circ}}{2}-\frac{\cos 50^{\circ}}{2} +\sqrt3\frac{\sin 50^{\circ}}{2}+\sqrt3\frac{\sin 30^{\circ}}{2}}{ \frac{\sin 50^{\circ}}{2}+\frac{\sin 30^{\circ}}{2}}$ $=\sqrt3$