If alpha = cos^{-1}left(frac{3}{5}right) and | vedclass

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Question

(A) Given $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.Thus

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$\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Vedclass · Answer

(A) Given $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{3/5} = \frac{4}{3}$.We are given $	an \beta = \frac{1}{3}$.Using the formula $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$,we get:$	an(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + (\frac{4}{3})(\frac{1}{3})} = \frac{1}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13}$.Now,if $	an(\alpha - \beta) = \frac{9}{13}$,then $\sin(\alpha - \beta) = \frac{9}{\sqrt{9^2 + 13^2}} = \frac{9}{\sqrt{81 + 169}} = \frac{9}{\sqrt{250}} = \frac{9}{5\sqrt{10}}$.Therefore,$\alpha - \beta = \sin^{-1}\left(\frac{9}{5\sqrt{10}}ight)$.

If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ and $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\alpha - \beta$ is equal to

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