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Question
Find the approximate values of : cot–1 (0.999)
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Solution
Let f(x) = cot–1 x
∴ f'(x) = `d/dx(cot^-1x) = (-1)/(1 + x^2)`
Take a = and h = – 0.001
Then f(a) = f(1) = cot–11 = `pi/(4)`
and f'(a) = f'(1) = `(-1)/(1 + 1^2) = (-1)/(2)`
The formula for appromation is
f(a + h) ≑ f(a) + h.f'(a)
∴ cot–1 (0.999)
= f(0.999)
= f(1 – 0.001)
≑ f(1) – (0.001).f'(1)
≑ `pi/(4) - (0001).((-1)/2)`
= `pi/(4) + 0.005`
∴ cot–1 (0.999) ≑ `pi/(4) + 0.0005`.
Remark: The answer can also be given as :
cot–1 (0.999) ≑ `(3.1416)/(4) + 0.0005`
≑ 0.7854 + 0.0005
= 0.7859.
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