Trigonometry
posted by Dee .
Find the exact value of tan(ab)
sin a = 4/5, 3pi/2<a<pi;
tan b = sqrt2, pi/2<b<pi
identity used is:
tan(ab)=(tan atan b)/1+tan a tan b
simplify answer using radicals.
(a is alpha, b is beta)

Both the sine and cosine curves are the same for
3pi/2<a<pi as they are for π/2<a<π (2nd quad)
so if sina = 4/5, then cosa = 3/5
and tana = 4/3
then tan(ab)
= (tana  tanb)/(1 + tanatanb)
= ((4/3)  (√2))/( 1 + (4/3)(√2))
= (√2  4/3)/(1 + 4√2/3)
= (3√2  4)/(3 + 4√2)
I don't know if you have to rationalize that, if you do carefully multiply top and bottom by (3  4√2) 
I put that in but it said the answer was
3625ã2/23
any idea how that works? 
That is exactly what my answer works out to if you rationalize it.
I had suggested to do that. 
8 cot(A) − 8/
1 + tan(−A)