The smallest and the largest values of {tan ^ | vedclass

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(B) Let $f(x) = 	an^{-1}\left(\frac{1-x}{1+x}ight)$.Using the property $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we ca

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$0, \frac{\pi}{4}$

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(B) Let $f(x) = 	an^{-1}\left(\frac{1-x}{1+x}ight)$.Using the property $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we can write:$f(x) = 	an^{-1}(1) - 	an^{-1}(x) = \frac{\pi}{4} - 	an^{-1}(x)$.Given the interval $0 \le x \le 1$,we know that $0 \le 	an^{-1}(x) \le \frac{\pi}{4}$.Multiplying by $-1$,we get $-\frac{\pi}{4} \le -	an^{-1}(x) \le 0$.Adding $\frac{\pi}{4}$ to all parts,we get $\frac{\pi}{4} - \frac{\pi}{4} \le \frac{\pi}{4} - 	an^{-1}(x) \le \frac{\pi}{4} - 0$.Thus,$0 \le f(x) \le \frac{\pi}{4}$.The smallest value is $0$ and the largest value is $\frac{\pi}{4}$.

The smallest and the largest values of ${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$ for $0 \le x \le 1$ are

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