The approximate value of tan ^{-1}(0.999) is | vedclass

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(B) Let $f(x) = 	an^{-1} x$. Then $f'(x) = \frac{1}{1+x^2}$. We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$. Here,let $a = 1

Vedclass · Accepted Answer

$0.7849$

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(B) Let $f(x) = 	an^{-1} x$. Then $f'(x) = \frac{1}{1+x^2}$. We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$. Here,let $a = 1$ and $h = -0.001$,so $a+h = 0.999$. $f(a) = 	an^{-1}(1) = \frac{\pi}{4}$. $f'(a) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5$. Substituting these values: $	an^{-1}(0.999) \approx \frac{\pi}{4} + (-0.001)(0.5)$. $	an^{-1}(0.999) \approx \frac{3.1415}{4} - 0.0005$. $	an^{-1}(0.999) \approx 0.785375 - 0.0005$. $	an^{-1}(0.999) \approx 0.784875$. Rounding to four decimal places,we get $0.7849$.

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )

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