Help with exact value of: $\tan (\sin^{-1}(-1/2) - \tan^{-1}(3/4))$ Ok, so I used the tan formula of difference of angles, and so far I've got to:
I found that $$\tan \alpha = \frac{-1/2}{\sqrt{3}/2} = \pm \frac{\sqrt3}{3}$$
so
$$\tan \left(\sin^{-1}\left(\frac{-1}{2}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{\sqrt3/3 - 3/4}{1+(\sqrt3/3)(3/4)}$$
I went a little past this but keep running into trouble. Can someone show me a step by step way to solve this from beginning to end so I can see how this is done? Also note it's not a proof, and I can't use exact values in form of decimals. It needs to be rationalized square-roots and fractions. Much appreciated
 A: Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have
$$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$
since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image.
Also,
$$\frac{\sin b}{\cos b}=\frac{3}{4} \Rightarrow \frac{9}{16}\cos^2 b+\cos^2b=1 \Rightarrow \cos b=\frac{4}{5} \Rightarrow \sin b=\frac{3}{5}$$
since $\tan^{-1}x$ has $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as image.
We then have
$$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$
$$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
A: You got to:
$$\tan\left[\sin^{-1}\frac{-1}2-\tan^{-1}\frac34\right]=\frac{\displaystyle \frac{\sqrt3}3-\frac34}{\displaystyle 1+\frac{\sqrt3}3\cdot\frac34}$$
And you need simplification:
Remember $\displaystyle\frac ab-\frac cd=\frac{ad-bc}{bc}$.  You will get the same denominator above and bellow the big fraction, so you will be able to drop it.
