Help with exact value of: $\tan (\sin^{-1}(-1/2) - \tan^{-1}(3/4))$
Andrew Mclaughlin 
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
$\endgroup$ 32 Answers
$\begingroup$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 $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$ $$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
$\endgroup$ 1 $\begingroup$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.
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