Approximate value of `tan^(-1)(0.999)` is

Options

A. `(pi)/4-0.005` radian
B. `(pi)/2-0.003` radia
C. `(pi)/3-0.002` radian
D. `(pi)/4-0.0005` radian

Answer

D

Explanation

Let `f(x)=tan^(-1)x`
`:.f'(x)=1/(1+x^(2))`
Take `a=1` and `h=-0.001`
`:'f(a+b)=f(a)+f'(a)h`
`:.tan^(-1)(a+h)=tan^(-1)a+1/(1+a^(2))/h`……………i
`:.tan^(-1)(0.999)=(pi)/4+1/(1+1)(-0.001)`
`=(pi)/4-0.001/2=(pi)/4=0.005` radian