Show that : ^ -1 (4 5 )+ ^ -1 (12 13 )= ^ -1 (33 65 )

Question

Show that : $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)$

✓

Answer

Let $a=\cos ^{-1}\left(\frac{4}{5}\right)$ and $b=\cos ^{-1}\left(\frac{12}{13}\right)$
Let $a=\cos ^{-1}\left(\frac{4}{5}\right)$
$\cos a =\frac{4}{5}$
We know that
$\sin^2a = 1 - \cos^2a $
$\sin a=\sqrt{1-\cos ^2 a}$
$=\sqrt{1-\left(\frac{4}{5}\right)^2}=\sqrt{1-\frac{16}{25}}$
$=\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
Let $b=\cos ^{-1}\left(\frac{12}{13}\right)$
$\cos b=\frac{12}{13}$
We know that
$\sin ^2 b=1-\cos ^2 b$
$\sin b=\sqrt{1-\cos ^2 b}$
$=\sqrt{1-\left(\frac{12}{13}\right)^2}=\sqrt{1-\frac{144}{169}}$
$=\sqrt{\frac{169-144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
We know that
$\cos (a+b) = \cos$ a $\cos b - \sin a \sin b$
Putting values
$\cos a=\frac{4}{5}, \sin a=\frac{3}{5}$
$\ \cos b=\frac{12}{13}, \sin b=\frac{5}{13}$
$\cos (a+b)=\frac{4}{5} \times \frac{12}{13} \times \frac{3}{5} \times \frac{5}{13}$
$=\frac{48}{65}-\frac{3}{13}$
$=\frac{48-15}{65}$
$=\frac{33}{65}$
$\therefore \cos (a+b)=\frac{33}{65}$
$a+b=\cos ^{-1}\left(\frac{33}{65}\right)$
$\cos ^{-1} \frac{4}{5}+\cos ^{-1}\left(\frac{12}{15}\right)=\cos ^{-1}\left(\frac{33}{65}\right)$
Hence $\text{LH.S = R.H.S}$
Hence proved.

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