4 tan^{-1} frac{1}{5} - tan^{-1} frac{1}{239} | vedclass

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(D) We know that $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$.First,calculate $2 	an^{-1} \frac{1}{5}$:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2(1

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) We know that $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$.First,calculate $2 	an^{-1} \frac{1}{5}$:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2(1/5)}{1-(1/5)^2} = 	an^{-1} \frac{2/5}{24/25} = 	an^{-1} \frac{5}{12}$.Now,calculate $4 	an^{-1} \frac{1}{5} = 2(2 	an^{-1} \frac{1}{5}) = 2 	an^{-1} \frac{5}{12}$:$2 	an^{-1} \frac{5}{12} = 	an^{-1} \frac{2(5/12)}{1-(5/12)^2} = 	an^{-1} \frac{10/12}{1-25/144} = 	an^{-1} \frac{5/6}{119/144} = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \frac{120}{119}$.Finally,subtract $	an^{-1} \frac{1}{239}$:$	an^{-1} \frac{120}{119} - 	an^{-1} \frac{1}{239} = 	an^{-1} \left( \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} 	imes \frac{1}{239}} ight)$,using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$.$= 	an^{-1} \left( \frac{28680 - 119}{28441 + 120} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1} (1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ is equal to

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