Answer
$\dfrac{\sqrt 6+\sqrt 2}{4}$
Work Step by Step
Here, we have $\sin [\cos^{-1} (\dfrac{1}{2}) +\tan^{-1} 1)=\sin(60^{\circ}+45^{\circ})$
Use formula $\sin (p+q)=\sin p\cos q +\cos p \sin q$
Thus, we have $\sin(60^{\circ}+45^{\circ})=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ} $
This gives:
$=(\dfrac{\sqrt 3}{2})(\dfrac{\sqrt 2}{2})+(\dfrac{1}{2})(\dfrac{\sqrt 2}{2})$
or, $=\dfrac{\sqrt 6+\sqrt 2}{4}$
