Trigonometric Functions — Maths STD 12 Science — Question

Let $\cos ^{-1} \frac{12}{13}=x$
$\therefore \cos x=\frac{12}{13}$
$\therefore \sin x=\frac{5}{13}$
and let $\sin ^{-1} \frac{3}{5}=y$
$\sin y=\frac{3}{5}$
$\therefore \cos y=\frac{4}{5}$
∴ using sin (x + y) = sin x cos y + cos x sin y
$=\frac{5}{13} \times \frac{4}{5}+\frac{12}{13} \times \frac{3}{5}$
$=\frac{20+36}{13 \times 5}$
$=\frac{56}{65}$
$\therefore x+y=\sin ^{-1} \frac{56}{65}$
$\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$
Hence proved.