If cos ^{-1}left(frac{5}{13}right)+cos ^{-1}l | vedclass

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(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$.Given $\cos ^{-1}\left(\frac{5}{13}\ri

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$\frac{-33}{65}$

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(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$.Given $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$.Here $A = \frac{5}{13}$ and $B = \frac{3}{5}$.Then $\sqrt{1-A^2} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.And $\sqrt{1-B^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.Substituting these values into the formula:$\cos ^{-1} x = \cos ^{-1} \left( \frac{5}{13} \cdot \frac{3}{5} - \frac{12}{13} \cdot \frac{4}{5} \right)$.$\cos ^{-1} x = \cos ^{-1} \left( \frac{15}{65} - \frac{48}{65} \right)$.$\cos ^{-1} x = \cos ^{-1} \left( \frac{15-48}{65} \right)$.$\cos ^{-1} x = \cos ^{-1} \left( \frac{-33}{65} \right)$.Therefore, $x = \frac{-33}{65}$.

If $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$, then $x$ is equal to

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