# Show that $\sin^{-1}{\frac{5}{13}}+\cos^{-1}{\frac{3}{5}}=\tan^{-1}{\frac{63}{16}}$

Toolbox:
• $$sin^{-1}x=tan^{-1}\bigg( \frac{x}{\sqrt{1-x^2}} \bigg)$$
• $$cos^{-1}x=tan^{-1} \bigg( \frac{\sqrt{1-x^2}}{x} \bigg)$$
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy},\:\:\:xy<1$$
L.H.S.=
By taking $$x=\frac{5}{13},\:\frac{x}{\sqrt{1-x^2}}=\frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}$$
By substituting in the above formula of $$sin^{-1}x$$ we get
$$sin^{-1}\frac{5}{13}=tan^{-1}\frac{5}{12}$$
By taking $$x=\frac{3}{5},\:\frac{\sqrt{1-x^2}}{x}=\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}=\frac{4}{5}.\frac{5}{3}=\frac{4}{3}$$
By substituting in the above formula of $$cos^{-1}x$$ we get
$$cos^{-1}\frac{3}{5}=tan^{-1}\frac{4}{3}$$
By substituting the values, the given expression becomes
$$sin^{-1}\frac{5}{13}+cos^{-1}\frac{3}{5}=tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3}$$
By taking $$x=\frac{5}{12},\:and\:y=\frac{4}{3}$$
$$\frac{x+y}{1-xy}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}=\frac{63}{36}.\frac{36}{16}=\frac{63}{16}$$
$$\Rightarrow\:tan^{-1}\bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}\bigg)=tan^{-1}\frac{63}{16}$$
=R.H.S.

edited Mar 7, 2013