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Question
Find the approximate value of tan−1 (1.002).
[Given: π = 3.1416]
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Solution
x = 1. 002
= 1 + 0.002
x = a + b
a = 1, b = 0.002
f(x) = tan−1 (x)
∴ f′(x) = `1/(1+x^2)`
∴ f'(1) = `1/(1 + 1) = 1/2`
We know that
f(a + h) ≑ f(a) + hf' (a)
Taking a = 1, h = 0.002
f(1 + 0.002) ≑ f(1) + (0.002)f'(1) ...(1)
Now f(x) = tan−1 x
∴ f(1) = tan−1 (1)
= `pi/4`
From (1)
f(1.002) ≑ `pi/4 + (0.002) (1/2)`
≑ `pi/4 + 0.001`
≑ `3.1416/4 + 0.001`
≑ 0.7854 + 0.001
≑ 0.7864
∴ tan−1 ≑ 0.7864
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