# Prove that:

Question:

Prove that:

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

Solution:

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

Formula Used: $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \times \sqrt{1-y^{2}}\right)$

Proof:

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

$=\cos ^{-1}\left(\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \times \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right)$

$=\cos ^{-1}\left(\frac{48}{65}-\sqrt{1-\frac{16}{25}} \times \sqrt{\left.1-\frac{144}{169}\right)}\right.$

$=\cos ^{-1}\left(\frac{48}{65}-\left(\sqrt{\frac{25-16}{25}} \times \sqrt{\frac{169-144}{169}}\right)\right)$

$=\cos ^{-1}\left(\frac{48}{65}-\left(\sqrt{\frac{9}{25}} \times \sqrt{\frac{25}{169}}\right)\right)$

$=\cos ^{-1}\left(\frac{48}{65}-\frac{3}{13}\right)$

$=\cos ^{-1}\left(\frac{48-15}{65}\right)$

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

$=$ RHS

Therefore, LHS = RHS

Hence proved.