Question

Show that:
$2{\sin }^{-1}\left(\frac{3}{5}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}$

Answer 1

$2{\sin }^{-1}\left(\frac{3}{5}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}$
Using formula, $2{\sin }^{-1}x={\cos }^{-1}(1-2x^2)$
$2{\sin }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}[1-2{\left(\frac{3}{5}\right)}^2]$
$={\cos }^{-1}\left[\frac{25-18}{25}\right]={\cos }^{-1}\frac{7}{25}$
Let, $x={\cos }^{-1}\frac{7}{25}$
$\cos x=\frac{7}{25}$
$\tan x=\frac{\sqrt{{\left(25\right)}^2-7^2}}{7}=\frac{24}{7}$
$x={\tan }^{-1}\left(\frac{24}{7}\right)$
$\therefore 2{\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{24}{7}\right)$
$\therefore$ LHS $={\tan }^{-1}\left(\frac{24}{7}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)$
${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$
$={\tan }^{-1}\left[\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times \frac{17}{31}}\right]$
$={\tan }^{-1}\left[\frac{24\times 31-7\times 17}{7\times 31+17\times 24}\right]$
$={\tan }^{-1}\left[\frac{744-119}{217+408}\right]={\tan }^{-1}\left[\frac{625}{625}\right]$
$={\tan }^{-1}\left(1\right)$
$=\frac{\pi }{4}$