If $\alpha ={\cos }^{-1}\left(\frac{3}{5}\right),\beta ={\tan }^{-1}\left(\frac{1}{3}\right)$  ,  $0<\alpha$   , $\beta <\frac{\pi }{2}$  , then  $\alpha -\beta$ = ________

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

$\alpha ={\cos }^{-1}\left(\frac{3}{5}\right)\Rightarrow \cos \alpha =\frac{3}{5}$

$\beta ={\tan }^{-1}\left(\frac{1}{3}\right)\Rightarrow \tan \beta =\frac{1}{3}$

We have  $\tan \alpha =\frac{4}{3}$ ,   $\tan \beta =\frac{1}{3}$  $\Rightarrow \tan (\alpha -\beta )=\frac{9}{13}$

 $\sin (\alpha -\beta )=\frac{9}{5\sqrt{10}}$

$\Rightarrow \alpha -\beta ={\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$