4 tan^{-1} frac{1}{5} - tan^{-1} frac{1}{70} | vedclass

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(C) हम सूत्र $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ का उपयोग करेंगे।सबसे पहले,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \fr

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$\frac{\pi}{4}$

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(C) हम सूत्र $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ का उपयोग करेंगे।सबसे पहले,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \frac{1}{5} ight) = 2 	an^{-1} \left( \frac{2/5}{1 - 1/25} ight) = 2 	an^{-1} \left( \frac{2/5}{24/25} ight) = 2 	an^{-1} \left( \frac{5}{12} ight)$.अब,$2 	an^{-1} \frac{5}{12} = 	an^{-1} \left( \frac{2(5/12)}{1 - (5/12)^2} ight) = 	an^{-1} \left( \frac{5/6}{1 - 25/144} ight) = 	an^{-1} \left( \frac{5/6}{119/144} ight) = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \left( \frac{120}{119} ight)$.आगे,हम $	an^{-1} \frac{1}{99} - 	an^{-1} \frac{1}{70} = 	an^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} ight) = 	an^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} ight) = 	an^{-1} \left( \frac{-29}{6931} ight) = -	an^{-1} \left( \frac{1}{239} ight)$ की गणना करते हैं।अंत में,$	an^{-1} \frac{120}{119} - 	an^{-1} \frac{1}{239} = 	an^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} ight) = 	an^{-1} \left( \frac{(28680 - 119) / 28441}{(28441 + 120) / 28441} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1}(1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = $

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