Question

Find the approximate value of tan⁻¹ (1.002).
[Given: π = 3.1416]

Answer 1

x = 1. 002

= 1 + 0.002

x = a + b

a = 1, b = 0.002

f(x) = tan⁻¹ (x)

∴ f′(x) = `1/(1+x^2)​`

∴ f'(1) = `1/(1 + 1) = 1/2`

We know that

f(a + h) ≑ f(a) + hf' (a)

Taking a = 1, h = 0.002

f(1 + 0.002) ≑ f(1) + (0.002)f'(1)    ...(1)

Now f(x) = tan⁻¹ x

∴ f(1) = tan⁻¹ (1)

= `pi/4`

From (1)

f(1.002) ≑ `pi/4 + (0.002) (1/2)`

≑ `pi/4 + 0.001`

≑ `3.1416/4 + 0.001`

≑ 0.7854 + 0.001

≑ 0.7864

∴ tan⁻¹ ≑ 0.7864