Question

Prove the following;

$2si{n}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{24}{7}$

Answer 1

Given, $2si{n}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{24}{7}$
LHS $=2si{n}^{-1}\frac{3}{5}=si{n}^{-1}[2\times \frac{3}{5}\sqrt{1-{\left(\frac{3}{5}\right)}^2}][\because 2si{n}^{-1}y=si{n}^{-1}(2y\sqrt{1-y^2}\left)\right]=si{n}^{-1}[2\times \frac{3}{5}\times \frac{4}{5}]=si{n}^{-1}\left(\frac{24}{25}\right)=ta{n}^{-1}\left[\frac{\frac{24}{25}}{\sqrt{1-{\left(\frac{24}{25}\right)}^2}}\right](\because si{n}^{-1}y=ta{n}^{-1}\frac{y}{\sqrt{1-y^2}})ta{n}^{-1}\left[\frac{\frac{24}{25}}{\sqrt{1-\frac{576}{625}}}\right]=ta{n}^{-1}[\frac{24}{25}\times \frac{25}{7}]=ta{n}^{-1}\left[\frac{24}{7}\right]$=RHS.