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

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(B) हम जानते हैं कि $2 	an ^{-1}(x) = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.इसलिए,$2 	an ^{-1} \left( \frac{1}{5} ight) = 	an ^{-1} \left(

Vedclass · Accepted Answer

$-\frac{7}{17}$

Vedclass · Answer

(B) हम जानते हैं कि $2 	an ^{-1}(x) = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.इसलिए,$2 	an ^{-1} \left( \frac{1}{5} ight) = 	an ^{-1} \left( \frac{2 	imes \frac{1}{5}}{1 - (\frac{1}{5})^2} ight) = 	an ^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} ight) = 	an ^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} ight) = 	an ^{-1} \left( \frac{2}{5} 	imes \frac{25}{24} ight) = 	an ^{-1} \left( \frac{5}{12} ight)$.अब,व्यंजक $	an \left[ 	an ^{-1} \left( \frac{5}{12} ight) - 	an ^{-1} (1) ight]$ हो जाता है।सूत्र $	an ^{-1}(x) - 	an ^{-1}(y) = 	an ^{-1} \left( \frac{x-y}{1+xy} ight)$ का उपयोग करने पर:$	an ^{-1} \left( \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} 	imes 1} ight) = 	an ^{-1} \left( \frac{-\frac{7}{12}}{\frac{17}{12}} ight) = 	an ^{-1} \left( -\frac{7}{17} ight)$.अंत में,$	an \left[ 	an ^{-1} \left( -\frac{7}{17} ight) ight] = -\frac{7}{17}$.

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

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