Question

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

Answer 1

Given, $ta{n}^{-1}\frac{63}{16}=si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}$
$RHS=si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}$
Let $si{n}^{-1}\left(\frac{5}{13}\right)=x\Rightarrow sinx=\frac{5}{13}\because 1-si{n}^{2}x=co{s}^{2}x\Rightarrow 1-{\left(\frac{5}{13}\right)}^2=co{s}^{2}x\Rightarrow \frac{169-25}{169}=co{s}^{2}x\Rightarrow \frac{144}{169}=co{s}^{2}x\Rightarrow \frac{12}{13}=cosx\therefore tanx=\frac{sinx}{cosx}\Rightarrow tanx=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}\Rightarrow x=ta{n}^{-1}\frac{5}{12}$

Let $co{s}^{-1}\frac{3}{5}=y\Rightarrow cosy=\frac{3}{5}\because 1-co{s}^{2}y=si{n}^{2}y\Rightarrow 1-{\left(\frac{3}{5}\right)}^2=si{n}^{2}y\Rightarrow \frac{25-9}{25}=si{n}^{2}y\Rightarrow \frac{16}{25}=si{n}^{2}y=\Rightarrow \frac{4}{5}=siny\therefore tany=\frac{siny}{cosy}\Rightarrow tany=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\Rightarrow y=ta{n}^{-1}\frac{4}{3}$
$\therefore$ Given equation becomes $ta{n}^{-1}\frac{63}{16}=x+y\Rightarrow ta{n}^{-1}\frac{63}{16}=ta{n}^{-1}\left(\frac{5}{12}\right)+ta{n}^{-1}\left(\frac{4}{3}\right)\therefore RHS=ta{n}^{-1}\left(\frac{5}{12}\right)+ta{n}^{-1}\left(\frac{4}{3}\right)=ta{n}^{-1}\left[\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}\right][\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)]=ta{n}^{-1}\left[\frac{\frac{15+48}{12\times 3}}{\frac{12\times 3-20}{12\times 3}}\right]=ta{n}^{-1}[\frac{63}{36}\times \frac{36}{36-20}]=ta{n}^{-1}\left[\frac{63}{16}\right]=LHS\therefore LHS=RHS$