Question

Show that:
${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=\pi$.

Answer 1

Let $x={\sin }^{-1}\left(\frac{12}{13}\right)$, then $\sin x=\frac{12}{13}$.

$\cos x=\sqrt{1-{\sin }^{2}x}$

$\cos x=\sqrt{1-{\left(\frac{12}{13}\right)}^2}$

$\cos x=\frac{5}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$\tan x=\frac{\frac{12}{13}}{\frac{5}{13}}$

$\tan x=\frac{12}{5}$

$x={\tan }^{-1}\left(\frac{12}{5}\right)$

So, ${\sin }^{-1}\left(\frac{12}{13}\right)={\tan }^{-1}\left(\frac{12}{5}\right)$

Similarly,${\cos }^{-1}\left(\frac{4}{5}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$

Substitute these values in the left hand side of ${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=\pi$.

$={\tan }^{-1}\left(\frac{12}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$

As $\frac{12}{5}\times \frac{3}{4}=\frac{9}{5}>1$, so, ${\tan }^{-1}A+{\tan }^{-1}B=\pi +{\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$.

$=\pi +{\tan }^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\times \frac{3}{4}}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$

$=\pi +{\tan }^{-1}\left(-\frac{63}{16}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$

$=\pi -{\tan }^{-1}\left(\frac{63}{16}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$

$=\pi$

$=RHS$

Hence LHS=RHS