Question

Prove that : ${\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{7}{25}={\cos }^{-1}\frac{253}{325}$

Answer 1

We have to prove: ${\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{7}{25}={\cos }^{-1}\frac{253}{325}$

Let $\alpha ={\sin }^{-1}\frac{5}{13},\beta ={\sin }^{-1}\frac{7}{25}$

$\Rightarrow \sin \alpha =\frac{5}{13},\sin \beta =\frac{7}{25}$

We know that, ${\cos }^2\alpha =1-{\sin }^2\alpha$

$\Rightarrow \cos \alpha =\sqrt{1-{\sin }^2\alpha }$

We have $\sin \alpha =\frac{5}{13}$

$\Rightarrow \cos \alpha =\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13}$

Similarly we have $\sin \beta =\frac{7}{25}$

$\Rightarrow \cos \beta =\sqrt{1-{\left(\frac{7}{25}\right)}^2}=\frac{24}{25}$

We have $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

$\cos (\alpha +\beta )=\frac{12}{13}\times \frac{24}{25}-\frac{5}{13}\times \frac{7}{25}$ $=\frac{288-35}{325}=\frac{253}{325}$

$\Rightarrow \cos (\alpha +\beta )=\frac{253}{325}$

$\Rightarrow \alpha +\beta ={\cos }^{-1}\left(\frac{253}{325}\right)$

Therefore, ${\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{7}{25}={\cos }^{-1}\frac{253}{325}$

Hence, proved.