pi + left(sin^{-1} frac{4}{5} + sin^{-1} frac | vedclass

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(C) We use the identity $\sin^{-1} x + \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} + y\sqrt{1-x^2})$.First,calculate $\sin^{-1} \frac{4}{5} + \sin^{-1} \fr

Vedclass · Accepted Answer

$\frac{3\pi}{2}$

Vedclass · Answer

(C) We use the identity $\sin^{-1} x + \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} + y\sqrt{1-x^2})$.First,calculate $\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13}$:$= \sin^{-1} \left(\frac{4}{5} \sqrt{1 - (\frac{5}{13})^2} + \frac{5}{13} \sqrt{1 - (\frac{4}{5})^2}ight)$$= \sin^{-1} \left(\frac{4}{5} 	imes \frac{12}{13} + \frac{5}{13} 	imes \frac{3}{5}ight)$$= \sin^{-1} \left(\frac{48}{65} + \frac{15}{65}ight) = \sin^{-1} \frac{63}{65}$.Now,the expression becomes $\pi + \sin^{-1} \frac{63}{65} + \sin^{-1} \frac{16}{65}$.Let $\alpha = \sin^{-1} \frac{63}{65}$,then $\cos \alpha = \sqrt{1 - (\frac{63}{65})^2} = \sqrt{\frac{4225 - 3969}{4225}} = \sqrt{\frac{256}{4225}} = \frac{16}{65}$.Thus,$\sin^{-1} \frac{63}{65} = \cos^{-1} \frac{16}{65}$.Substituting this back: $\pi + \cos^{-1} \frac{16}{65} + \sin^{-1} \frac{16}{65}$.Since $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we get $\pi + \frac{\pi}{2} = \frac{3\pi}{2}$.

$\pi + \left(\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65}\right)$ is equal to

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