If cos^{-1} frac{3}{5} - sin^{-1} frac{4}{5} | vedclass

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(B) Given equation: $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$We know that $\sin^{-1} 	heta = \cos^{-1} \sqrt{1 - 	heta^2}$ for $

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$1$

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(B) Given equation: $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$We know that $\sin^{-1} 	heta = \cos^{-1} \sqrt{1 - 	heta^2}$ for $	heta \in [0, 1]$.Therefore,$\sin^{-1} \frac{4}{5} = \cos^{-1} \sqrt{1 - (\frac{4}{5})^2} = \cos^{-1} \sqrt{1 - \frac{16}{25}} = \cos^{-1} \sqrt{\frac{9}{25}} = \cos^{-1} \frac{3}{5}$.Substituting this back into the original equation:$\cos^{-1} \frac{3}{5} - \cos^{-1} \frac{3}{5} = \cos^{-1} x$$0 = \cos^{-1} x$$x = \cos(0) = 1$.

If $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$,then $x = $

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