Question

Prove: ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

Answer 1

Let ${\cos }^{-1}\frac{4}{5}=x\Rightarrow \cos x=\frac{4}{5}\Rightarrow \sin x=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5}$
$\therefore \tan x=\frac{3}{4}\Rightarrow x={\tan }^{-1}\frac{3}{4}$
$\therefore {\cos }^{-1}\frac{4}{5}={\tan }^{-1}\frac{3}{4}.....\left(1\right)$
Now, let ${\cos }^{-1}\frac{5}{12}=y\Rightarrow \cos y=\frac{12}{13}\Rightarrow \sin y=\frac{5}{13}$
$\therefore \tan y=\frac{12}{13}\Rightarrow y={\tan }^{-1}\frac{5}{12}$
$\therefore {\cos }^{-1}\frac{12}{13}={\tan }^{-1}\frac{5}{12}.....\left(2\right)$
Let ${\cos }^{-1}\frac{33}{65}=z\Rightarrow \cos z=\frac{33}{65}\Rightarrow \sin z=\frac{56}{65}$
$\therefore \tan z=\frac{56}{33}\Rightarrow z={\tan }^{-1}\frac{56}{33}$
$\therefore {\cos }^{-1}\frac{33}{65}={\tan }^{-1}\frac{56}{33}.....\left(3\right)$
Now,
L.H.S. ${=\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}$
$={\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}$ [using (1) and (2)]
$={\tan }^{-1}\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}},[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}]$
$={\tan }^{-1}\frac{36+20}{48-15}$
$={\tan }^{-1}\frac{56}{33}$ [By 3]
$=$ R.H.S.