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

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Question

(D) આપણે જાણીએ છીએ કે $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$.પ્રથમ,$2 	an^{-1} \frac{1}{5}$ ની ગણતરી કરો:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$.પ્રથમ,$2 	an^{-1} \frac{1}{5}$ ની ગણતરી કરો:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2(1/5)}{1-(1/5)^2} = 	an^{-1} \frac{2/5}{24/25} = 	an^{-1} \frac{5}{12}$.હવે,$4 	an^{-1} \frac{1}{5} = 2(2 	an^{-1} \frac{1}{5}) = 2 	an^{-1} \frac{5}{12}$ ની ગણતરી કરો:$2 	an^{-1} \frac{5}{12} = 	an^{-1} \frac{2(5/12)}{1-(5/12)^2} = 	an^{-1} \frac{10/12}{1-25/144} = 	an^{-1} \frac{5/6}{119/144} = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \frac{120}{119}$.અંતે,$	an^{-1} \frac{1}{239}$ બાદ કરો:$	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)$,સૂત્ર $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$ નો ઉપયોગ કરતા.$= 	an^{-1} \left( \frac{28680 - 119}{28441 + 120} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1} (1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ ની કિંમત શોધો.

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