Easy Trigonometry

Squaring $\sin \theta$ = $\frac{1}{3}$ , we have ( $\sin \theta$ )^2 = 1/9.

Using the identity ( $\sin \theta$ )^2 = 1- ( $\cos \theta$ )^2, we have ( $\cos \theta$ )^2= 8/9.

Simplifying, we get $\cos \theta$ = $\frac{2\sqrt{2}}{3}$ (positive because of the condition $\cos \theta$ >0).

Since $\tan \theta$ = $\frac{\sin \theta}{\cos \theta}$ , we have $\tan \theta$ = $\frac{\sqrt{2}}{4}$ .

Thus, we have a= 2 and b= 4 .

Finally, we have $\boxed{a+b= 6}$ .

from pythagoras theorem we find tdat the other vertical is square root of 8 ,so tanθ equal 1 over square root of 8 so a=2 and b=4