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# What does $2\sin^{-1} \frac{3}{5}$ reduce to

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A)
• $$sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}$$
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} |x| < 1$$
Let $x =$ $$\large\frac{3}{5}$$, $$\rightarrow \large \frac{x}{\sqrt{1-x^2}}=\large\frac{\large\frac{3}{5}}{\sqrt{1-\large\frac{9}{25}}}=\large\frac{3}{4}$$
$$\Rightarrow\:sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}$$
$$\Rightarrow 2sin^{-1}\large\frac{3}{5}$$ $$=2 \tan^{-1}\large\frac{3}{4}$$
Given $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$, we need to evaluate $$2tan^{-1}\large\frac{3}{4}$$
Let $x =\large\frac{3}{4} \rightarrow \large \large\frac{2x}{1-x^2}=\large\frac{2.\frac{3}{4}}{1-\large\frac{9}{16}}=\large\frac{6}{4}.\large\frac{16}{7}=\large\frac{24}{7}$
$$\Rightarrow\:2tan^{-1}\large\frac{3}{4}$$$$= tan^{-1}\large\frac{24}{7}$$