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Question
Find the approximate values of : tan–1(0.999)
Solution
Let f(x) = tan–1x
Then f'(x) = `d/dx(tan^-1x) = (1)/(1 + x^2)`
Take a = 1 and h = – 0.001
Then f(a) = f(1) = tan–11 = `pi/(4)`
and f'(a) = f'(1) = `(1)/(1 + 1^2) = (1)/(2)`
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ tan–1 (0.999)
= f(0.999)
= f(1 – 0.001)
≑ f(1) – (0.001).f'(1)
≑ `pi/(4) - 0.001 xx (1)/(2)`
= `pi/(4) - 0.0005`
∴ tan–1 (0999) ≑ `pi/(4) - 0.0005`.
Remark: The answer can also be given as :
tan–1 (0.999) ≑ `(3.1416)/(4) - 0.0005`
≑ 0.7854 – 0.0005
= 0.7849.
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