Prove that $$\tan^{-1} \large\frac{63}{16}=\sin^{-1}\bigg( \large\frac{5}{13} \bigg) + \cos^{-1} \bigg( \large\frac{3}{5} \bigg).$$

This is Q.No.7 of Misc. chapter 2

Toolbox:
• $$sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$
• $$cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}$$
• $$tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}$$
Given R.H.S. = $sin^{-1} \frac{5}{13}+cos^{-1} \frac {3}{5}$
We know that $$sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$
By taking $x=\frac{5}{13} \rightarrow$ $$\large \frac{x}{\sqrt{1-x^2}}$$=$$\Large \frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}$$$$=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}$$
$$\Rightarrow\:sin^{-1}\frac{5}{13}=tan^{-1}\frac{5}{12}$$
We know that $$cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}$$
By taking $x=$$$\frac{3}{5} \rightarrow \large \frac{\sqrt{1-x^2}}{x}=\Large \frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}$$$$=\frac{4}{5}.\frac{5}{3}=\frac{3}{4}$$
$$\Rightarrow\:cos^{-1}\frac{3}{5}=tan^{-1}\frac{4}{3}$$
$$\Rightarrow sin^{-1} \frac{5}{13}+cos^{-1}\frac{3}{5}$$ $$= tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3}$$
$$\frac{x+y}{1-xy}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}=\frac{63}{36}.\frac{36}{16}=\frac{63}{16}$$
$$= tan^{-1} \bigg[ \frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{20}{36}} \bigg]$$
$$= tan^{-1}\frac{63}{16}$$
=L.H.S
We know that $$tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}$$
By taking $$x=\frac{5}{12}\:and\:y=\frac{4}{3}\:in\:tan^{-1}x+tan^{-1}y$$
$x+y = \frac{5}{12}+\frac{4}{3} = \frac{63}{36}$
$1 - xy = 1 - \frac{5}{12} \times \frac{4}{3} = \frac{16}{36}$
$\large \frac{x+y}{1-xy} = \frac{63}{36}.\frac{36}{16}=\frac{63}{16}$
$\Rightarrow tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3} =$ $$tan^{-1}\frac{63}{16}$$ = L.H.S.
edited Mar 18, 2013