tan left[ 2tan^{-1}left( frac{1}{5} right) - | vedclass

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Question

(D) આપણે સૂત્ર $2	an^{-1}(x) = 	an^{-1}\left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$2	an^{-1}\left( \frac{1}{5} ight)$ ની ગણતરી કરો:$2

Vedclass · Accepted Answer

$-\frac{7}{17}$

Vedclass · Answer

(D) આપણે સૂત્ર $2	an^{-1}(x) = 	an^{-1}\left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$2	an^{-1}\left( \frac{1}{5} ight)$ ની ગણતરી કરો:$2	an^{-1}\left( \frac{1}{5} ight) = 	an^{-1}\left( \frac{2(1/5)}{1-(1/5)^2} ight) = 	an^{-1}\left( \frac{2/5}{1-1/25} ight) = 	an^{-1}\left( \frac{2/5}{24/25} ight) = 	an^{-1}\left( \frac{2}{5} 	imes \frac{25}{24} ight) = 	an^{-1}\left( \frac{5}{12} ight)$.હવે,આ કિંમતને પદાવલિમાં મૂકો:$	an \left[ 	an^{-1}\left( \frac{5}{12} ight) - \frac{\pi}{4} ight] = 	an \left[ 	an^{-1}\left( \frac{5}{12} ight) - 	an^{-1}(1) ight]$.સૂત્ર $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}\left( \frac{x-y}{1+xy} ight)$ નો ઉપયોગ કરતા:$	an \left[ 	an^{-1}\left( \frac{5/12 - 1}{1 + (5/12)(1)} ight) ight] = \frac{5/12 - 1}{1 + 5/12} = \frac{-7/12}{17/12} = -\frac{7}{17}$.

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

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