Petrolovujhm

2022-03-27

Help with exact value of:
$\mathrm{tan}\left({\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)-{\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)\right)$

Drake Huang

Let $a={\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)$ and $b={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)$. We have
${\mathrm{sin}}^{2}a+{\mathrm{cos}}^{2}a=1⇒\frac{1}{4}+{\mathrm{cos}}^{2}a=1⇒\mathrm{cos}a=\frac{\sqrt{3}}{2}$
$⇒\mathrm{tan}a=-\frac{\sqrt{3}}{3}$
since ${\mathrm{sin}}^{-1}x$ has $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ as image.
Also,
$\frac{\mathrm{sin}b}{\mathrm{cos}b}=\frac{3}{4}⇒\frac{9}{16}{\mathrm{cos}}^{2}b+{\mathrm{cos}}^{2}b=1⇒\mathrm{cos}b=\frac{4}{5}⇒\mathrm{sin}b=\frac{3}{5}$
since ${\mathrm{tan}}^{-1}x$ has $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ as image.
we then have
$\mathrm{tan}\left(a-b\right)=\frac{\mathrm{tan}a-\mathrm{tan}b}{1+\mathrm{tan}a\mathrm{tan}b}=\frac{-\frac{\sqrt{3}}{3}-\frac{3}{4}}{1-\frac{\sqrt{3}}{3}\frac{3}{4}}=-\frac{4\sqrt{3}+9}{12-3\sqrt{3}}$
$=-\frac{4\sqrt{3}+9}{12-3\sqrt{3}}\frac{12+3\sqrt{3}}{12+3\sqrt{3}}$
$=-\frac{108+48\sqrt{3}+27\sqrt{3}+36}{144-27}$
$=-\frac{144+75\sqrt{3}}{117}$
$=-\frac{48+25\sqrt{3}}{39}$

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