Question

Find the approximate value of ${\tan ^{ - 1}}\left( {1.001} \right)$
A. 0.6655
B. 0.9358
C. 0.5555
D. 0.7855

Answer 1

Hint: In order to solve this question we will use the property of differentiation of function which can be expressed as $f(x + \Delta x) = f\left( x \right) + \Delta f'\left( x \right)$ where $\left( {\Delta x < < < x} \right)$ and then by simplifying it we will get the answer.

Complete step-by-step answer:
Let $f\left( y \right) = {\tan ^{ - 1}}y$
Differentiating $f\left( y \right)$ w.r.t $y$ we have
As we know that \[\left[ {\dfrac{{d\left( {{{\tan }^{ - 1}}y} \right)}}{{dx}} = \dfrac{1}{{1 + {y^2}}}} \right]\]
$ \Rightarrow \dfrac{{df\left( y \right)}}{{dx}} = \dfrac{1}{{1 + {y^2}}}...................\left( 1 \right)$
Let $y = 1.001 = x + \Delta x$
Here,
$x = 1$
And $\Delta x = .001$
Substituting the values of x in above equations
Therefore, $f\left( x \right) = f\left( 1 \right) = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}{\text{ }}$
From equation (1)
Similarly, $f'\left( x \right) = f'\left( 1 \right) = \dfrac{1}{{1 + {1^2}}} = \dfrac{1}{2}$
Now, from Taylor series expansion
$
  f\left( y \right) = f\left( {x + \Delta x} \right) = f\left( x \right) + \Delta x.f'\left( x \right) \\
  {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {x + \Delta x} \right) = {\tan ^{ - 1}}x + \Delta x.\left( {\dfrac{1}{{1 + {x^2}}}} \right) \\
  \therefore {\tan ^{ - 1}}1.001 = {\tan ^{ - 1}}\left( {1 + 0.001} \right) = {\tan ^{ - 1}}1 + \left( {0.001} \right).{\tan ^{ - 1}}1 \\
   \Rightarrow {\tan ^{ - 1}}1.001 = \dfrac{\pi }{4} + 0.001\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow {\tan ^{ - 1}}1.001 = \dfrac{\pi }{4} + 0.0005 = 0.7855 \\
$
Hence, the approximate value of ${\tan ^{ - 1}}\left( {1.001} \right)$ is $0.7855$ and the correct option is “D”.

Note: In order to solve these types of questions, we must have a good concept of differentiation of trigonometric functions as well as inverse trigonometric functions. The above question is an application of differentiation. In the above question we used the Taylor series expansion to calculate the value of a given function. Differentiation can also be used to find the maxima and minima of a function.