Prove: $\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$ This is not a homework question, its from sl loney I'm just practicing.
To prove :
$$\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$$
So I changed all the angles to $\arctan$ which gives:
$$\arctan\left(\frac 34\right) - \arctan\left(\frac {12}{5}\right) = \arctan\left(\frac {16}{63}\right)$$
But the problem is after applying formula of $\arctan(X)-\arctan(Y)$ the lhs is negative and not equal to rhs? Is this because I have to add pi somewhere please help.

To specifically use $arctan$ formula:
Change all angles to $tan$ as you outlined, using Pythagoras...
$$\arctan(\frac34) - \arctan(\frac5{12}) = \arctan(\frac{16}{63})$$
Apply  $arctan$ formula 
$$\arctan(x) - \arctan(y) = \arctan(\frac{x-y}{1+xy})$$
$$\arctan(\frac{\frac34 - \frac5{12}}{1+\frac34 \frac5{12}}) = \arctan(\frac{16}{63})$$
Simplify to give
$$\arctan(\frac{16}{63}) = \arctan(\frac{16}{63})$$
$$LHS = RHS$$
