Answer
-0.08
Work Step by Step
For small $\Delta x$, $\Delta y\approx dy$.
That is, $\Delta y\approx f'(a) dx$.
$f'(a)=2\tan a \sec^{2} a=2\tan\frac{\pi}{4}\sec^{2}\frac{\pi}{4}$
$= 2\times1\times2=4$
$dx=-0.02$
$\implies \Delta y\approx 4\times-0.02=-0.08$