tan ^{-1} frac{3}{4} + tan ^{-1} frac{3}{5} - | vedclass

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Question

(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$ for $xy < 1$. First,evaluate $	an ^{-1} \frac{3}{4}

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(A) We use the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$ for $xy First,evaluate $	an ^{-1} \frac{3}{4} + 	an ^{-1} \frac{3}{5}$: $	an ^{-1} \left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} 	imes \frac{3}{5}} ight) = 	an ^{-1} \left( \frac{\frac{15+12}{20}}{1 - \frac{9}{20}} ight) = 	an ^{-1} \left( \frac{27/20}{11/20} ight) = 	an ^{-1} \frac{27}{11}$. Now,subtract $	an ^{-1} \frac{8}{19}$ using $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left( \frac{x-y}{1+xy} ight)$: $	an ^{-1} \frac{27}{11} - 	an ^{-1} \frac{8}{19} = 	an ^{-1} \left( \frac{\frac{27}{11} - \frac{8}{19}}{1 + \frac{27}{11} 	imes \frac{8}{19}} ight)$. Calculate the numerator: $\frac{27 	imes 19 - 8 	imes 11}{11 	imes 19} = \frac{513 - 88}{209} = \frac{425}{209}$. Calculate the denominator: $1 + \frac{216}{209} = \frac{209 + 216}{209} = \frac{425}{209}$. Thus,$	an ^{-1} \left( \frac{425/209}{425/209} ight) = 	an ^{-1}(1) = \frac{\pi }{4}$.

$\tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{3}{5} - \tan ^{-1} \frac{8}{19} = $

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