Question

Show that

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

Answer 1

L.H.S. $=2{\sin }^{-1}\left(\frac{3}{5}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)$

$={\sin }^{-1}(2\times \frac{3}{5}\sqrt{1-\frac{9}{25}})-{\tan }^{-1}\left(\frac{17}{31}\right)$

$={\sin }^{-1}(\frac{6}{5}\times \frac{4}{5})-{\tan }^{-1}\left(\frac{17}{31}\right)$

$={\sin }^{-1}\left(\frac{24}{25}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)$

$={\tan }^{-1}\left(\frac{24}{7}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)$

$={\tan }^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times \frac{17}{31}}\right)$

$={\tan }^{-1}\left(\frac{744-119}{217+408}\right)$

$={\tan }^{-1}\left(\frac{625}{625}\right)$

$={\tan }^{-1}\left(1\right)=\frac{\pi }{4}=$ R.H.S.