Show that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
(A) Let $\sin ^{-1} \frac{12}{13}=x$,$\cos ^{-1} \frac{4}{5}=y$,and $\tan ^{-1} \frac{63}{16}=z$.
Then $\sin x = \frac{12}{13}$,$\cos y = \frac{4}{5}$,and $\tan z = \frac{63}{16}$.
From these,we find $\cos x = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,$\sin y = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.
Thus,$\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{5/13} = \frac{12}{5}$ and $\tan y = \frac{\sin y}{\cos y} = \frac{3/5}{4/5} = \frac{3}{4}$.
Now,calculate $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - (\frac{12}{5} \times \frac{3}{4})} = \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} = \frac{63/20}{-16/20} = -\frac{63}{16}$.
Since $\tan(x+y) = -\frac{63}{16}$ and $\tan z = \frac{63}{16}$,we have $\tan(x+y) = -\tan z = \tan(\pi - z)$.
Since $x, y, z$ are in the first quadrant,$x+y+z = \pi$.
Then $\sin x = \frac{12}{13}$,$\cos y = \frac{4}{5}$,and $\tan z = \frac{63}{16}$.
From these,we find $\cos x = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,$\sin y = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.
Thus,$\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{5/13} = \frac{12}{5}$ and $\tan y = \frac{\sin y}{\cos y} = \frac{3/5}{4/5} = \frac{3}{4}$.
Now,calculate $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - (\frac{12}{5} \times \frac{3}{4})} = \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} = \frac{63/20}{-16/20} = -\frac{63}{16}$.
Since $\tan(x+y) = -\frac{63}{16}$ and $\tan z = \frac{63}{16}$,we have $\tan(x+y) = -\tan z = \tan(\pi - z)$.
Since $x, y, z$ are in the first quadrant,$x+y+z = \pi$.
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