# Prove that $2sin^{-1} \frac{3}{5} = tan^{-1} \frac{24}{7}$

Toolbox:
• $$sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}$$
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} |x| < 1$$
Given $2sin^{-1}\large \frac{3}{5} = tan^{-1} \large\frac{24}{7}$

L.H.S.: Let $x =$ $$\large\frac{3}{5}$$, $$\rightarrow \large \frac{x}{\sqrt{1-x^2}}=\large\frac{\large\frac{3}{5}}{\sqrt{1-\large\frac{9}{25}}}=\large\frac{3}{4}$$

$$\Rightarrow\:sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}$$
$$\Rightarrow 2sin^{-1}\large\frac{3}{5}=2tan^{-1}\large\frac{3}{4}$$
Given $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$, we need to evaluate $$2tan^{-1}\large\frac{3}{4}$$

Let $x =\large\frac{3}{4} \rightarrow \large \large\frac{2x}{1-x^2}=\large\frac{2.\frac{3}{4}}{1-\large\frac{9}{16}}=\large\frac{6}{4}.\large\frac{16}{7}=\large\frac{24}{7}$

$$\Rightarrow\:2tan^{-1}\large\frac{3}{4} = tan^{-1}\large\frac{24}{7} =$$ R.H.S.
answered Jul 9, 2014