Question

Prove that: ${\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\frac{56}{65}$

Answer 1

${\cos }^{-1}\frac{12}{13}$
$={\sin }^{-1}\sqrt{1-{\left(\frac{12}{13}\right)}^2}\left({\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right)$
$={\sin }^{-1}\sqrt{1-\frac{144}{169}}$
$={\sin }^{-1}\sqrt{\frac{25}{169}}$
$={\sin }^{-1}\frac{5}{13}$

$$\begin{gathered} \therefore {\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5} \\ ={\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{3}{5} \\ ={\sin }^{-1}\left(\frac{5}{13}\times \sqrt{1-{\left(\frac{3}{5}\right)}^2}+\frac{3}{5}\times \sqrt{1-{\left(\frac{5}{13}\right)}^2}\right)\left[{\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)\right] \end{gathered}$$
$$\begin{gathered} ={\sin }^{-1}\left(\frac{5}{13}\times \sqrt{1-\frac{9}{25}}+\frac{3}{5}\times \sqrt{1-\frac{25}{169}}\right) \\ ={\sin }^{-1}\left(\frac{5}{13}\times \sqrt{\frac{16}{25}}+\frac{3}{5}\times \sqrt{\frac{144}{169}}\right) \\ ={\sin }^{-1}\left(\frac{5}{13}\times \frac{4}{5}+\frac{3}{5}\times \frac{12}{13}\right) \end{gathered}$$
$$\begin{gathered} ={\sin }^{-1}\left(\frac{20}{65}+\frac{36}{65}\right) \\ ={\sin }^{-1}\frac{56}{65} \end{gathered}$$