Find tan(arcsin(4/6) + arccos(10/17))

Question

PLEASE show clear steps!!

If the answers ends up having square roots, can you simply it where it does not

Answer 1

Whenever you need to find some func(arcfunc(a/b)), draw the triangle which fits. In this case, let

sinA = 4/6

if you draw the triangle, you can see that tanA = 4/√20 = 2/√5

similarly, if cosB = 10/17, then tanB = √189/10 = 3√21/10

Now you just need to use the sum of tangents formula to find

tan(A+B) = (2/√5 + 3√21/10) / (1 - 2/√5 * 3√21/10) = -(289√5 + 135√21)/128

or maybe

(20√5 - 15√21) / (50 - 6√105)

or maybe just -9.8818