Question

Show that ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\cot }^{-1}\frac{63}{16}$ $=\frac{\pi }{2}$

Answer 1

${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\cot }^{-1}\frac{63}{16}{\sin }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{63}{16}{\sin }^{-1}(\frac{12}{13}\times \frac{4}{5}+\frac{3}{5}\times \frac{5}{13})+{\cot }^{-1}\frac{63}{16}{\sin }^{-1}(\frac{48}{65}+\frac{15}{65})+{\sin }^{-1}\frac{16}{65}{\sin }^{-1}\frac{63}{65}+{\cos }^{-1}\frac{63}{65}=\pi /2$