Decent Trigonometry!

$\large\tan \theta=\dfrac 34\implies \cos \theta=\dfrac 45$

$\large{\begin{aligned}\cos\left(\dfrac{\theta}2\right)=&\sqrt{\dfrac{1+\cos \theta}2}\\=&\sqrt{\dfrac{1+\frac 45}2}=\dfrac3{\sqrt{10}}\end{aligned}}$

$\huge\therefore~ 3+10=\boxed{\color{#D61F06}{13}}$

$(\text{Since }\cos \theta>0,\cos\theta/2>0)$

Since it is given that $\theta$ lies in the first quadrant, we can form a triangle with sides of lengths $3,4,5$ units (or, in general, $3a,4a,5a$ units). Suppose that $\angle A=\theta°$ and $\angle B=90°$ . Thus $AB=4$ , $BC=3$ and $AC=5$ .

Draw an angle bisector from vertex $A$ on to the side $BC$ . Call the point where the angle bisector touches the side $BC$ as $D$ .

Then by angle bisector theorem, $\frac{AC}{AB} = \frac{CD}{DB}$ . Hence length of the line segment $DB=\frac{4}{3}$ .

Now $\triangle ABD$ contains the angle $\large{\frac{\theta°}{2}}$ .

Since $\triangle ABD$ is a right triangle, ${AB}^2 + {BD}^2 = {AD}^2$ . Hence $AD=\frac{4\sqrt{10}}{3}$ .

Finally, $\cos\angle DAB = \large{\cos\frac{\theta°}{2}} = \frac{AB}{AD} = \frac{3}{\sqrt{10}}$ .

$\begin{aligned} \color{#3D99F6}{\tan \theta} & = \frac 34 \quad \quad \small \color{#3D99F6}{\text{Let }t = \tan \frac \theta 2} \\ \color{#3D99F6}{\frac{2t}{1-t^2}} & = \frac 34 \\ \implies 3t^2 + 8t-3 & = 0 \\ (3t-1)(t+3) & = 0 \\ \implies \tan \frac \theta 2 & = \frac 13 \quad \quad \small \color{#3D99F6}{\text{for } 0 \le \theta \le \frac \pi 2 \implies \tan \frac \theta 2 > 0} \\ \implies \cos \frac \theta 2 & = \frac 3{\sqrt{1^2+3^2}} = \frac 3{\sqrt 10} \end{aligned}$

$\implies x + y = 3 + 10 = \boxed{13}$

Take a representation $\theta = 2x$ . We should find $\cos x$ given $\tan 2x$ . Since $\tan 2x\ = 3/4$ , it can be derived from the triangle trigonometic rule that $\cos 2x = 4/5$ . After that use trigonometric formula $\cos 2x = 2cos^2x - 1$ to find $\cos x$ , which is $3/\sqrt 10$ . Then the answer is $3 + 10 = 13$ .