Let's see
both those things in the brackets are angles.
so the question becomes sin(A+B)=sinACosB+cosAsinB
Now
It needs to be remembered that
the range for both inverse sine and inverse tan is from -pi/2 to pi/2 inclusive
A=sin^-1(1/4) so A is in the 1st quad sinA=1/4 and cosA=sqrt(15 )/4
B=tan^-1(-5) so B is in the 4th quad tanB=-5 and sinB=-5/sqrt(26) and cosB=+1/sqrt(26)
I got these ratios by drawing the right angled triangles for A and B and finig the third side use Pythagoras' theorm.
$$\\sin(A+B)\\\\
=sinACosB+cosAsinB\\\\
=\frac{1}{4}\times\frac{1}{\sqrt{26}}+\frac{\sqrt{15}}{4}\times\frac{-5}{\sqrt26}\\\\
=\frac{1}{4\sqrt{26}}+\frac{-5\sqrt{15}}{4\sqrt{26}}\\\\
=\frac{1-5\sqrt{15}}{4\sqrt{26}}\\\\
=\frac{\sqrt{26}\left(1-5\sqrt{15}\right)}{104}\\\\$$
(I have edited this a little as I made a mistake)
If you want me to explain anything just ask but think about it a bit first :)
Here are the triangles that I used.
