Question

Prove that ${\cos }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}={\tan }^{-1}\frac{56}{33}$.

Answer 1

${\cos }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}$

$={\tan }^{-1}\left(\frac{5}{12}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12}\times \frac{3}{4}}\right)$

$={\tan }^{-1}\left(\frac{\frac{20+36}{48}}{1-\frac{15}{48}}\right)$

${\tan }^{-1}\left(\frac{56}{33}\right)$ hence proved