I'm going to substitute "a" for alpha and "b" for beta just to make it easier to type out the solution to the problem. It's important to notice that sin(a) and tan(b) are both positive.
Here's the sum identity we need to solve his problem:
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
Let's assume a and b are both in quadrant I where sin(a) and tan(b) are positive. (There's an additional answer if a is in quadrant II and b is in quadrant III which we'll get to later.)
We're given sin(a)=3/5 and tan(b)=5/12, but we need to know what tan(a) is in order to solve the problem so let's start there.
Recall that the sine of any angle is y/r.
In other words, sin(a)=y/r
Also recall that the tangent of any angle is y/x. In other words, tan(a)=y/x
We need to know what x is, and we can find it using the Pythagorean theorem: x2+y2=r2
sin(a)=3/5, so y=3 and r=5.
Plugging these values into the the Pythagorean theorem, we get:
x2+32=52
x2+9=25
x2=16
x=4 (x could also be -4 which we'll get to later)
Now that we know what x is, we can find tan(a):
tan(a)=y/x
tan(a)=3/4
So we have tan(a)=3/4 and we were given tan(b)=5/12.
Now we can plug these values into the sum identity and simplify:
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
=(3/4+5/12)/(1-(3/4)(5/12))
=(7/6)/(1-(5/16))
=(7/6)/(11/16)
=56/33
So one answer: tan(a+b)=56/33
Now since sin(a) is also positive in quadrant II and tan(b) is also positive in quadrant III, x would be -4 in that case. This will lead us to the second solution.
When x=-4, tan(a)=-3/4.
Let's plug this into our identity:
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
=(-3/4+5/12)/(1-(-3/4)(5/12))
=(-1/3)/(1-(-5/16))
=(-1/3)/(1+5/16)
=(-1/3)/(21/16)
=-16/63
So that's our second answer: tan(a+b)=-16/63
I hope you found this helpful!
