Question

Prove that ${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{119}{169}$

Answer 1

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

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

We get-

${\cos }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{12}{13}$

Now, $2{\cos }^{-1}x={\cos }^{-1}(2x^2-1)$

$⟹2{\cos }^{-1}\frac{12}{13}={\cos }^{-1}(2{\left(\frac{12}{13}\right)}^2-1)$

$={\cos }^{-1}\left(\frac{288-169}{169}\right)$

$={\cos }^{-1}\frac{119}{169}$