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Find the approximate values of : tan–1 (1.001) - Mathematics and Statistics

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प्रश्न

Find the approximate values of : tan–1 (1.001)

योग

उत्तर

Let f(x) = tan–1x

∴ f'(x) = `d/dx(tan^-1x) = (1)/(1 + x^2)`

Take a = 1 and h = 0.001

Then f(a) = f(1) = tan–11 = `pi/(4)`

and f'(a) = f'(1) = `(1)/(1 + 1^2) = (1)/(2)`
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ tan–11 (1.001)
= f(1.001)
= f(1 + 0.001)
= f(1) + (0.001).f'(1)

≑ `pi/(4) + (0.001) xx (1)/(2)`

= `pi/(4) + 0.0005`

∴ tan–1 (1.001) ≑ `pi/(4) + 0.0005`.
Remark: the answer can also be given as :
tan–1 (1.001) ≑ f(1) + (0.001).f'(1)

≑ `pi/(4) + (0.001) xx (1)/(2)`

≑ `(3.1416)/(4) + 0.0005`

≑ 0.7854 + 0.0005
= 0.7859.

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अध्याय 2: Applications of Derivatives - Exercise 2.2 [पृष्ठ ७५]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 2 Applications of Derivatives
Exercise 2.2 | Q 3.3 | पृष्ठ ७५

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