# Tan[arctan(2/3) + arccos(8/17)]

1. Jul 29, 2011

### jrjack

1. The problem statement, all variables and given/known data

tan[arctan(2/3)+arccos(8/17)]

2. Relevant equations

tan(u+v)=$\frac{\tan u+\tan v}{1-\tan u\tan v}$

3. The attempt at a solution

After drawing 2 triangles for arctan and arccos (both in Quadrant 1), and inserting them into the addition formula for tan, I get:
$$\frac{\frac{2}{3}+\frac{15}{8}}{1-\frac{2}{3}(\frac{15}{8})}$$
is that right?
if so, then maybe my math is off...
I got the numerator reduced to 23/8
and denominator of -1/4
for a total of -23/3 ...but thats not correct.

2. Jul 29, 2011

### Dick

Re: tan[arctan(2/3)+arccos(8/17)]

I think your math with the fractions is off. 2/3+15/8 is not equal to 23/8. Can you show why you think it is?

3. Jul 29, 2011

### jrjack

Re: tan[arctan(2/3)+arccos(8/17)]

Sorry, mis-typed...I meant 23/12

2/3 =16/24
15/8=30/24******Found my problem, this should be 45/24

Thanks for looking at this, I knew it had to be something simple...I need to slow down.