Question

Prove, $co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}$

Answer 1

Given, $co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}$
Let $co{s}^{-1}\frac{12}{13}=\theta \Rightarrow cos\theta =\frac{12}{13}\therefore sin\theta =\sqrt{1-{\left(\frac{12}{13}\right)}^2}[\because sin\theta =\sqrt{1-co{s}^2\theta }]\Rightarrow sin\theta =\sqrt{\frac{25}{169}}=\frac{5}{13}\Rightarrow \theta =si{n}^{-1}\frac{5}{13}$
Now, $LHS=co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\left(\frac{5}{13}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right)+\frac{3}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}[\because si{n}^{-1}x+si{n}^{-1}y=si{n}^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}\left)\right]=si{n}^{-1}(\frac{5}{13}\sqrt{\frac{16}{25}}+\frac{3}{5}\sqrt{\frac{144}{169}})=si{n}^{-1}(\frac{5}{13}\times \frac{4}{5}+\frac{3}{5}\times \frac{12}{13})=si{n}^{-1}\left(\frac{56}{65}\right)=RHS$
Hence proved.