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