Find the approximate value of ^ -1 (0.99), Given that 3.1416. - Vidyadip

Question

Find the approximate value of $\tan ^{-1}(0.99)$, Given that $\pi \doteqdot 3.1416$.

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Answer

Let $f(x)=\tan ^{-1} x$
Differentiate w.r.t. $x$.
$f^{\prime}(x)=\frac{1}{1+x^2}$
Let $a=1, h=-0.01$
For $x=a=1$, from (I) we get
$f(a)=f(1)=\tan ^{-1}(1)=\frac{\pi}{4}$
For $x=a=1$, from (II) we get
$f^{\prime}(a)=f^{\prime}(1)=\frac{1}{1+1^2}=0.5$
We have, $f(a+h) \doteqdot f(a)+h f^{\prime}(a)$
$f[(1)+(-0.01)] \doteqdot f(1)+(-0.01) \cdot f^{\prime}(1)$
$f(0.99) \doteqdot \frac{\pi}{4}-(0.01) \cdot(0.5) \ldots[$ From
(III) and (IV)]
$\doteqdot \frac{\pi}{4}-0.005$
$\doteqdot \frac{3.1416}{4}-0.005$
$\doteqdot 0.7854-0.005=0.7804$
$\therefore f(0.99)=\tan ^{-1}(0.99) \doteqdot 0.7804$

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