Question

Prove: ${\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\frac{56}{65}$

Answer 1

Let ${\sin }^{-1}\frac{3}{5}=x$. Then, $\sin x=\frac{3}{5}\Rightarrow \cos x=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

$\therefore \tan x=\frac{3}{4}\Rightarrow x={\tan }^{-1}\frac{3}{4}$

$\therefore {\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$ ....(1)

Now, let ${\cos }^{-1}\frac{12}{13}=y$. Then $\cos y=\frac{12}{13}=\sin y=\frac{5}{13}$

$\therefore \tan y=\frac{5}{12}\Rightarrow y={\tan }^{-1}\frac{5}{12}$

${\therefore \cos }^{-1}\frac{12}{13}={\tan }^{-1}\frac{5}{12}$ ......(2)

Let ${\sin }^{-1}\frac{56}{65}=z$. Then, $\sin z=\frac{56}{65}\Rightarrow \cos z=\frac{33}{65}$

$\therefore \tan z=\frac{56}{33}\Rightarrow z={\tan }^{-1}\frac{56}{33}$

$\therefore {\sin }^{-1}\frac{56}{65}={\tan }^{-1}\frac{56}{33}$ .......(3)

Now, we have:
L.H.S. $={\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}$

$={\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{3}{4}$ ...[using (1) and (2) ]

$={\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12}.\frac{3}{4}}\right)$ ...$[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}]$

$={\tan }^{-1}\frac{20+36}{48-15}$

$={\tan }^{-1}\frac{56}{33}$

$={\sin }^{-1}\frac{56}{65}=$ R.H.S. [using (3)]