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The approximate vale of ${Tan}^{- 1} (0.999)$ (upto 4 decimal places) is

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0.7852

0.7102

0.8124

0.7526

answer is A.

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Detailed Solution

$\begin{array}{rcl} x & = & 1 σx = - 0.001 \\ f (x) & = & \tan^{- 1} x \\ f (x + δ x) & = & \frac{1}{1 + x^{2}} δ x + f (x) \\ = & \frac{- 0.001}{2} + \frac{π}{4} \\ = & 0.785 - 0.0005 = 0.7845 \end{array}$

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