Show that

${\cos }^{-1}\frac{3}{5}+{cosec}^{-1}\sqrt{5}={\tan }^{-1}\frac{11}{2}$

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

$4\left({\cot }^{-1}3+{cosec}^{-1}\sqrt{5}\right)=\pi$

$2{\tan }^{-1}\frac{1+x}{1-x}-{\cos }^{-1}\frac{1-x^2}{1+x^2}=\frac{\pi }{2}$

If $\left|x\right|\le 1$, then $\sin \left\{{cosec}^{-1}\cot \left({\tan }^{-1}x\right)\right\}=x$

$\sin \left[{\cot }^{-1}\left\{\cos \left({\tan }^{-1}x\right)\right\}\right]=\sqrt{\frac{x^2+1}{x^2+2}}$, if $x>0$.

$\cos \left[{\tan }^{-1}\left\{\sin \left({\cot }^{-1}x\right)\right\}\right]=\sqrt{\frac{x^2+1}{x^2+2}}$

If $x\ne 1$, then $\tan \left(\frac{1}{2}{\sin }^{-1}\frac{2x}{1+x^2}+\frac{1}{2}{\cos }^{-1}\frac{1-x^2}{1+x^2}\right)=\frac{2x}{1-x^2}$

${\tan }^{-1}\frac{a-b}{1+ab}+{\tan }^{-1}\frac{b-c}{1+bc}+{\tan }^{-1}\frac{c-d}{1+cd}={\tan }^{-1}a-{\tan }^{-1}d$