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

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(C) આપણે સૂત્ર $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીએ છીએ.પ્રથમ,$2 	an ^{-1} \frac{1}{5} = 	an ^{-1} \left( \fr

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$\frac{\pi}{4}$

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(C) આપણે સૂત્ર $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીએ છીએ.પ્રથમ,$2 	an ^{-1} \frac{1}{5} = 	an ^{-1} \left( \frac{2/5}{1-1/25} ight) = 	an ^{-1} \left( \frac{2/5}{24/25} ight) = 	an ^{-1} \frac{5}{12}$ ગણો.ત્યારબાદ,$4 	an ^{-1} \frac{1}{5} = 2 	an ^{-1} \frac{5}{12} = 	an ^{-1} \left( \frac{2(5/12)}{1-(5/12)^2} ight) = 	an ^{-1} \left( \frac{5/6}{1-25/144} ight) = 	an ^{-1} \left( \frac{5/6}{119/144} ight) = 	an ^{-1} \frac{120}{119}$.હવે,$-	an ^{-1} \frac{1}{70} + 	an ^{-1} \frac{1}{99} = 	an ^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} ight) = 	an ^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} ight) = 	an ^{-1} \left( \frac{-29}{6931} ight) = -	an ^{-1} \frac{1}{239}$.અંતે,$	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)/(119 	imes 239)}{(28441+120)/(119 	imes 239)} 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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