Question

Find the approximate values of : cot^–1 (0.999)

Answer 1

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.