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Find the approximate values of : tan (45° 40'), given that 1° = 0.0175°.

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प्रश्न

Find the approximate values of : tan (45° 40'), given that 1° = 0.0175°.

बेरीज
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उत्तर

Let f(x) = tan x

Then f'(x) = `d/dx(tanx) = sec^2x`

Take a = 45°

= `pi/(4)`
and
h = 40'

= `(40/60 xx 0.0175)^c`

= 0.01167c

Then f(a) = `f(pi/4)`

= `tan  pi/(4)`
= 1
and
f'(a) = `f'(pi/4)`

= `sec^2  pi/(4)`

= `(sqrt(2))^2`
= 2
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ tan(45° 40')
= f(45° 40')

= `f(pi/4 + 0.01167)`

≑ `f(pi/4) + (0.01167).f'(pi/4)`

≑ 1 + 0.01167 x 2 
= 1 + 0.02334
= 1.02334
∴ tan (45° 40') ≑ 1.02334.

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

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 2 Applications of Derivatives
Exercise 2.2 | Q 2.4 | पृष्ठ ७५

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