Question

Prove:

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

Answer 1

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

Let ${\sin }^{-1}\frac{3}{5}=a$

$\Rightarrow \sin a=\frac{3}{5}$

$\Rightarrow {\sin }^{2}a=\frac{9}{25}$

$\Rightarrow {\cos }^{2}a=1-{\sin }^{2}a=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$

$\Rightarrow \cos a=\frac{4}{5}$

$\therefore \tan a=\frac{\sin a}{\cos a}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

$\Rightarrow a={\tan }^{-1}\frac{3}{4}$

$\therefore {\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$

Hence

$2{\sin }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{17}{31}$$=2{\tan }^{-1}\frac{3}{4}-{\tan }^{-1}\frac{17}{31}$

We know that,

$2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

L.H.S$={\tan }^{-1}\left(\frac{2\times \frac{3}{4}}{1-{\left(\frac{3}{4}\right)}^2}\right)-{\tan }^{-1}\frac{17}{31}$

L.H.S$={\tan }^{-1}\left(\frac{\frac{6}{4}}{1-\frac{9}{16}}\right)-{\tan }^{-1}\frac{17}{31}$

L.H.S$={\tan }^{-1}\left(\frac{\frac{6}{4}}{\frac{16-9}{16}}\right)-{\tan }^{-1}\frac{17}{31}$

L.H.S$={\tan }^{-1}\frac{24}{7}-{\tan }^{-1}\frac{17}{31}$

We know that ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$

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

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

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

L.H.S$={\tan }^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right)$

L.H.S$={\tan }^{-1}1=\frac{\pi }{4}=$R.H.S

Hence proved.