Question

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

Answer 1

Let ${\sin }^{-1}\frac{5}{13}=\theta$

then $\sin \theta =\frac{5}{13}$

$\because \cos \theta =\sqrt{1-{\sin }^2\theta }$

$\therefore \cos \theta =\sqrt{1-{\left(\frac{5}{13}\right)}^2}$

$\Rightarrow \cos \theta =\frac{12}{13}$

$\Rightarrow \theta ={\cos }^{-1}\frac{12}{13}$

$\Rightarrow {\sin }^{-1}\frac{5}{13}={\cos }^{-1}\frac{12}{13}$

L.H.S $={\cos }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}$

$={\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}$

$={\cos }^{-1}\left[\frac{4}{5}\times \frac{12}{13}-\sqrt{1-{\left(\frac{4}{5}\right)}^2}\sqrt{1-{\left(\frac{12}{13}\right)}^2}\right]$ $[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2}\left)\right]$

$={\cos }^{-1}\left[\frac{48}{65}-\frac{3}{5}\times \frac{5}{13}\right]$

$={\cos }^{-1}\left[\frac{48}{65}-\frac{15}{65}\right]$

$={\cos }^{-1}\left[\frac{48-15}{65}\right]$

$={\cos }^{-1}\frac{33}{65}$

$=$ R.H.S.