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.

How exactly did you convert to arctan? Careful:
$$\arccos\left(\frac {12}{13}\right) = \arctan\left(\frac {5}{12}\right)
\ne \arctan\left(\frac {12}{5}\right)$$
Draw a right triangle with hypotenuse of length 13, adjacent side (from an angle $\alpha$) with length 12 and opposite side with length 5; then $\cos\alpha = 12/13$ and $\tan\alpha = 5/12$.

Perhaps easier: take the sine of both sides in the original equation:
$$\sin\left(\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right)\right) = \sin\left( \arcsin\left(\frac {16}{65}\right)\right)$$
The RHS is $16/65$ and simplify the LHS with $\sin(a-b)=\sin a \cos b - \cos a \sin b$ to get:
$$\frac{3}{5}\frac{12}{13}-\sqrt{1-\frac{9}{25}}\sqrt{1-\frac{144}{169} } = \frac{3}{5}\frac{12}{13}-\frac{4}{5}\frac{5}{13}= \cdots$$
