The value of tan left[2 tan ^{-1} frac{1}{5}- | vedclass

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(C) Let $	heta = 2 	an^{-1} \frac{1}{5}$.Using the formula $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$,we have:$	heta = 	an^{-1} \

Vedclass · Accepted Answer

$-\frac{7}{17}$

Vedclass · Answer

(C) Let $	heta = 2 	an^{-1} \frac{1}{5}$.Using the formula $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$,we have:$	heta = 	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)$.Now,we need to evaluate $	an \left( 	heta - \frac{\pi}{4} ight)$.Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,where $A = 	an^{-1} \frac{5}{12}$ and $B = \frac{\pi}{4}$:$	an \left( 	an^{-1} \frac{5}{12} - \frac{\pi}{4} ight) = \frac{\frac{5}{12} - 	an \frac{\pi}{4}}{1 + \frac{5}{12} 	an \frac{\pi}{4}} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}(1)} = \frac{\frac{5-12}{12}}{\frac{12+5}{12}} = \frac{-7}{17}$.

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

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