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

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(D) Let the given expression be $S = \sin \left(	an ^{-1} \frac{4}{5}+	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}-	an ^{-1} \frac{1}{7}ight)$.Re

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(D) Let the given expression be $S = \sin \left(	an ^{-1} \frac{4}{5}+	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}-	an ^{-1} \frac{1}{7}ight)$.Rearranging the terms, we get $S = \sin \left(\left(	an ^{-1} \frac{4}{5}-	an ^{-1} \frac{1}{7}ight) + \left(	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}ight)ight)$.Let $p = 	an ^{-1} \frac{4}{5}-	an ^{-1} \frac{1}{7}$. Using the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left(\frac{x-y}{1+xy}ight)$, we have $p = 	an ^{-1} \left(\frac{\frac{4}{5}-\frac{1}{7}}{1+\frac{4}{5} \cdot \frac{1}{7}}ight) = 	an ^{-1} \left(\frac{\frac{28-5}{35}}{\frac{35+4}{35}}ight) = 	an ^{-1} \left(\frac{23}{39}ight)$.Let $q = 	an ^{-1} \frac{4}{3}+	an ^{-1} \frac{1}{9}$. Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$, we have $q = 	an ^{-1} \left(\frac{\frac{4}{3}+\frac{1}{9}}{1-\frac{4}{3} \cdot \frac{1}{9}}ight) = 	an ^{-1} \left(\frac{\frac{12+1}{9}}{\frac{27-4}{27}}ight) = 	an ^{-1} \left(\frac{13}{9} \cdot \frac{27}{23}ight) = 	an ^{-1} \left(\frac{39}{23}ight)$.Since $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$, we have $q = \cot ^{-1} \left(\frac{23}{39}ight) = \frac{\pi}{2} - 	an ^{-1} \left(\frac{23}{39}ight)$.Thus, $p+q = 	an ^{-1} \left(\frac{23}{39}ight) + \frac{\pi}{2} - 	an ^{-1} \left(\frac{23}{39}ight) = \frac{\pi}{2}$.Therefore, $S = \sin \left(\frac{\pi}{2}ight) = 1$.

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

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