Question

Show that $si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{63}{16}.$

Answer 1

We have, $si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{63}{16}.$

$Letsi{n}^{-1}\frac{5}{13}=x\Rightarrow sinx=\frac{5}{13}andco{s}^{2}x=1-si{n}^{2}x=1-\frac{25}{169}=\frac{144}{169}\Rightarrow cosx=\sqrt{\frac{144}{169}}=\frac{12}{13}\therefore tanx=\frac{sinx}{cosx}=\frac{5/13}{12/13}=\frac{5}{12}\cdots \left(ii\right)\Rightarrow tanx=5/12\cdots \left(iii\right)$

$Again,letco{s}^{-1}\frac{3}{5}=y\Rightarrow cosy=\frac{3}{5}\therefore siny=\sqrt{1-co{s}^{2}y}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}siny=\sqrt{\frac{16}{25}}=\frac{4}{5}\Rightarrow tany=\frac{siny}{cosy}=\frac{4/5}{3/5}=\frac{4}{3}$
We know that,
$tan(x+y)=\frac{tanx+tany}{1-tanx.tany}$

$\Rightarrow tan(x+y)=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}\Rightarrow tan(x+y)=\frac{\frac{15+48}{36}}{\frac{36-20}{36}}\Rightarrow tan(x+y)=\frac{63/36}{16/36}\Rightarrow tan(x+y)=\frac{63}{16}\Rightarrow x+y=ta{n}^{-1}\frac{63}{16}\Rightarrow ta{n}^{-1}\frac{5}{12}+ta{n}^{-1}\frac{4}{3}=ta{n}^{-1}\frac{63}{16}Henceproved.$