Alan Smith

2022-01-17

Finding the value of ${\mathrm{sin}}^{-1}\frac{12}{13}+{\mathrm{cos}}^{-1}\frac{45}{+}{\mathrm{tan}}^{-1}\frac{63}{13}$

RizerMix

Expert

Actually
${\mathrm{tan}}^{-1}\frac{12}{5}+{\mathrm{tan}}^{-1}\frac{3}{4}=\phantom{\rule{0ex}{0ex}}=\pi +{\mathrm{tan}}^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot \frac{3}{4}}\right)$
We can notice that
$\frac{\pi }{2}<{\mathrm{tan}}^{-1}\frac{12}{5}+{\mathrm{tan}}^{-1}\frac{3}{4}<\pi$
I am going to prove that
NSK
if
Proof:
$\mathrm{tan}\left(\alpha +\beta -\pi \right)=\frac{\mathrm{tan}\alpha +\mathrm{tan}\left(\beta -\pi \right)}{1-\mathrm{tan}\alpha \mathrm{tan}\left(\beta -\pi \right)}=\frac{\mathrm{tan}\alpha +\mathrm{tan}\beta }{1-\mathrm{tan}\alpha \cdot \mathrm{tan}\beta }$
As $|\alpha +\beta -\pi |<\frac{\pi }{2}$ and the function tangent is invertible in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ it follows that
$\alpha +\beta -\pi ={\mathrm{tan}}^{-1}\left(\frac{\mathrm{tan}\alpha +\mathrm{tan}\beta }{1-\mathrm{tan}\alpha \cdot \mathrm{tan}\beta }\right)$ , therefore:
$\phantom{\rule{0ex}{0ex}}\alpha +\beta =\pi +{\mathrm{tan}}^{-1}\left(\frac{\mathrm{tan}\alpha +\mathrm{tan}\beta }{1-\mathrm{tan}\alpha \cdot \mathrm{tan}\beta }\right)$

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