\(\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}+\cos ^{-1} \frac{63}{65}=\frac{\pi}{2}\)
माना कि \(\cos ^{-1} \frac{3}{5}=y_1 \cos ^{-1} \frac{12}{13}=y_2\)
\(y_1+y_2 =\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13} \)
\(y_1+y_2 =\cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right) \)
\(=\cos ^{-1}\left(\frac{36}{65}-\frac{4 \times 5}{5 \times 13}\right)=\cos ^1\left(\frac{36}{65}-\frac{20}{65}\right)=\cos ^{-1} \frac{16}{65}\)
पुन: \(\cos ^{-1} \frac{16}{65}=\theta \Rightarrow \cos \theta=\frac{16}{65}\)
\(\sin \theta=\sqrt{1-\left(\frac{16}{65}\right)^2}=\sqrt{\frac{3969}{65 \times 65}}=\frac{63}{65} \)
\( \therefore \theta=\sin ^{-1} \frac{63}{65}\)
अत: \(y_1+y_2+\cos ^{-1} \frac{63}{65}\)
\(\sec ^{-1} \frac{63}{65}+\cos ^{-1} \frac{63}{15}=\frac{\pi}{2}\) Proved.
