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MHT CET · Maths · Application of Derivatives

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

  1. A 0.7843
  2. B 0.7849
  3. C 0.7847
  4. D 0.7851
Verified Solution

Answer & Solution

Correct Answer

(B) 0.7849

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \text { Let } \mathrm{f}(x)=\tan ^{-1} x \\
\therefore \quad & \mathrm{f}^{\prime}(x)=\frac{1}{1+x^2}
\end{aligned}\)
Here, \(\mathrm{a}=1\) and \(\mathrm{h}=-0.001\)
\(\begin{aligned}
\therefore \quad f(a+h) \approx f(a) & +h f^{\prime}(a) \\
\therefore \quad \tan ^{-1}(0.999) & \approx \frac{\pi}{4}+\frac{1}{1+1}(-0.001) \\
& \approx \frac{\pi}{4}-\frac{0.001}{2} \\
& \approx \frac{\pi}{4}-0.0005 \\
& \approx \frac{3.1415}{4}-0.0005 \\
& \approx 0.7849
\end{aligned}\)