Right option is (b) \(\frac{2π}{3}\)
Easy explanation: Using the formula sin^-1x+sin^-1y=sin^-1\({x \sqrt{1-y^2}+y \sqrt{1-x^2}}\), we get
sin^-1(\(\frac{3}{5}\))+sin^-1(\(\frac{4}{5}\))=sin^-1\(\Big\{ \frac{3}{5} \sqrt{1-(\frac{4}{5})^2}+\frac{4}{5} \sqrt{1-(\frac{3}{5})^2} \Big\}\)
=sin^-1(\(\frac{3}{5}\)×\(\frac{3}{5}\)+\(\frac{4}{5}\)×\(\frac{4}{5}\))=sin^-1\((\frac{25}{25})=\frac{π}{2}\)
∴ sin^-1(\(\frac{3}{5}\))+sin^-1(\(\frac{4}{5}\))+cos^-1\((\frac{\sqrt{3}}{2})=\frac{π}{2}+\frac{π}{6}=\frac{3π+π}{6}=\frac{2π}{3}\).