Browse Questions

# 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