1.

Find the value of sin^-1⁡(\(\frac{3}{5}\))+sin^-1⁡(\(\frac{4}{5}\))+cos^-1⁡\((\frac{\sqrt{3}}{2})\).(a) \(\frac{π}{3}\)(b) \(\frac{2π}{3}\)(c) \(\frac{4π}{3}\)(d) \(\frac{π}{4}\)This question was posed to me in an online interview.The doubt is from Properties of Inverse Trigonometric Functions in chapter Inverse Trigonometric Functions of Mathematics – Class 12

Answer»

Right option is (B) \(\frac{2π}{3}\)

Easy explanation: USING the formula sin^-1⁡X+sin^-1⁡y=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⁡\((\frac{\sqrt{3}}{2})=\frac{π}{2}+\frac{π}{6}=\frac{3π+π}{6}=\frac{2π}{3}\).


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