# Inverse trig question?

Hello everyone I have a question about trig.

How would I solve the following.

$$\tan\left(2\arcsin(4/5)+\arccos(12/13)\right)=\frac{253}{204}$$

-
I'm guessing you meant $\tan\Big(2\arcsin(4/3)+\arccos(12/13)\Big)$ and not $\tan\Big(2\arcsin(4/3)\Big)+\arccos(12/13)$. – Michael Hardy Dec 13 '12 at 22:47
Solve? There are no variables. Or do you mean how does one show that the left-hand side equals the right-hand side? – JohnD Dec 13 '12 at 22:47
The expression $\arcsin(4/3)$ is problematic. The domain of the conventional $\arcsin$ function is $[-1,1]$, and $4/3$ is not in that interval. Is there any chance you meant $4/5$? – Michael Hardy Dec 13 '12 at 22:51
What is the formula for tan(x+y)? When you use is, you will expressions of the form tan(asin(...)) and tan(acos(...)). Look at a right triangle to see how to compute those expressions. – marty cohen Dec 13 '12 at 22:51
It's false. The LHS is complex. Notice that $4/3 > 1$. – Ayman Hourieh Dec 13 '12 at 22:51

$$\tan(x+y+z) = \frac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan x\tan z-\tan y\tan z}\tag{1}$$
You want $$\tan\left(\arcsin\frac45 + \arcsin\frac45 + \arccos \frac{12}{13}\right).$$ So there you have $x$, $y$, and $z$.
Draw the right triangle in which the "opposite" side is $4$ and the hypotenuse is $5$. By the Pythagorean theorem, the "adjacent" side is $3$. So $\arcsin\frac45=\arctan\frac43$, so $\tan x=\frac43$. Then draw on where the "adjacent" side is $12$ and the hypotenuse is $13$, and then find that the "opposite" side is $5$, so the tangent is $5/12$. Then you have $\tan z=\frac{5}{12}$. Plug those tangents into the right side of $(1)$ and simplify.
And recall that if you know $\tan(x+y) = \dfrac{\tan x+\tan y}{1-\tan x\tan y}$, then you can derive $(1)$ by saying $\tan(x+y+z) = \tan(\Big(x\Big)+\Big(y+z\Big))$, etc.