Answer
$2$
Work Step by Step
We have:
$\sin\dfrac{\pi}{2}=1$
We work on the expression $\tan\dfrac{19\pi}{4}$:
$\tan\dfrac{19\pi}{4}=\tan\left(4\pi+\dfrac{3\pi}{4}\right)$
$=\tan\dfrac{3\pi}{4}$ (because the period of tangent is $\pi$)
$=-\tan\dfrac{\pi}{4}$
(The reference angle of $\frac{3\pi}{4}$ is $\frac{\pi}{4 }$ and since $\frac{3\pi}{4}$ is in Quadrant II, the value of tangent is negative.)
$\tan{\frac{\pi}{4}}=-1$
Evaluate the given expression to obtain:
$\sin\dfrac{\pi}{2}-\tan\dfrac{19\pi}{4}=1-(-1)=2$
