Question

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

Answer 1

Let f(x) = cot^–1 x

∴ f'(x) = ${\displaystyle \frac{d}{dx}\left({\cot }^{-1}x\right)=\frac{-1}{1+x^2}}$

Take a =  and h = – 0.001

Then f(a) = f(1) = cot^–11 = ${\displaystyle \frac{\pi }{4}}$

and f'(a) = f'(1) = ${\displaystyle \frac{-1}{1+1^2}=\frac{-1}{2}}$
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)

≑ ${\displaystyle \frac{\pi }{4}-\left(0001\right).\left(\frac{-1}{2}\right)}$

= ${\displaystyle \frac{\pi }{4}+0.005}$

∴ cot^–1 (0.999) ≑ ${\displaystyle \frac{\pi }{4}+0.0005}$.
Remark: The answer can also be given as :

cot^–1 (0.999) ≑ ${\displaystyle \frac{3.1416}{4}+0.0005}$

≑ 0.7854 + 0.0005
= 0.7859.