Question

Show that ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\pi$

Answer 1

Let ${\sin }^{-1}\frac{12}{13}=x\Rightarrow \sin x=\frac{12}{13}$
Now we have
$\Rightarrow \cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-\frac{144}{169}}=\frac{5}{13}$
$\Rightarrow \tan x=\frac{\sin x}{\cos x}=\frac{12}{5}$

Let ${\cos }^{-1}\frac{4}{5}=y\Rightarrow \cos y=\frac{4}{5}$
$\Rightarrow \sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
$\Rightarrow \tan y=\frac{\sin y}{\cos y}=\frac{3}{4}$

Let ${\tan }^{-1}\frac{63}{16}=z\Rightarrow \tan z=\frac{63}{16}$

Now we have,
$\tan x=\frac{12}{5},\tan y=\frac{3}{4},\tan z=\frac{63}{16}$
$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$
$=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot \frac{3}{4}}=\frac{\frac{48+15}{20}}{\frac{20-36}{20}}=-\frac{63}{16}$
$\Rightarrow \tan (x+y)=-\tan z$
i.e., $\tan (x+y)=\tan (-z)\text{or}\tan (x+y)=\tan (\pi -z)$
Therefore $x+y=-z\text{or}x+y=\pi -z$
Since $x,y\text{and}z$ are positive, $x+y\ne -z$
$\Rightarrow x+y=\pi -z$
$\Rightarrow x+y+z=\pi$ or
${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\pi$