Question

Prove that $si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{77}{85}.$

Answer 1

We have, $si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{77}{85}$

$\therefore LHS=si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{8}{15}+ta{n}^{-1}\frac{3}{4}$

$Letsi{n}^{-1}\frac{8}{17}={\theta }_1\Rightarrow sin{\theta }_1=\frac{8}{17}\Rightarrow tan{\theta }_1=\frac{8}{15}\Rightarrow {\theta }_1=ta{n}^{-1}\frac{8}{15}andsi{n}^{-1}\frac{3}{5}={\theta }_2\Rightarrow sin{\theta }_2=\frac{3}{5}\Rightarrow tan{\theta }_2=\frac{3}{4}\Rightarrow {\theta }_2=ta{n}^{-1}\frac{3}{4}$

$=ta{n}^{-1}\left[\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}\right][\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)]=ta{n}^{-1}\left[\frac{\frac{32+45}{60}}{\frac{60-24}{60}}\right]=ta{n}^{-1}\left(\frac{77}{36}\right)$

$Let{\theta }_3=ta{n}^{-1}\frac{77}{36}\Rightarrow \tan {\theta }_3=\frac{77}{36}\Rightarrow sin{\theta }_3=\frac{77}{\sqrt{5929+1296}}=\frac{77}{85}\therefore {\theta }_3=si{n}^{-1}\frac{77}{85}=si{n}^{-1}\frac{77}{85}=RHSHenceproved.$

Alternate Method

To prove, $si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{8}{5}=si{n}^{-1}\frac{77}{85}$

$Letsi{n}^{-1}\frac{8}{17}=x\Rightarrow sinx=\frac{8}{17}\Rightarrow cosx=\sqrt{1-si{n}^{2}x}=\sqrt{1-{\left(\frac{8}{17}\right)}^2}=\sqrt{\frac{289-64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$

$Letsi{n}^{-1}\frac{3}{5}=y\Rightarrow siny=\frac{3}{5}\Rightarrow si{n}^{2}y=\frac{9}{25}\therefore co{s}^{2}y=1-\frac{9}{25}\Rightarrow co{s}^{2}y={\left(\frac{4}{5}\right)}^2\Rightarrow cosy=\frac{4}{5}Now,sin(x+y)=sinx.xosy+cosx.siny=\frac{8}{17}.\frac{4}{5}+\frac{15}{17}.\frac{3}{5}=\frac{32}{85}+\frac{45}{85}=\frac{77}{85}\Rightarrow (x+y)=si{n}^{-1}\left(\frac{77}{85}\right)\Rightarrow si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{77}{85}$