Question

How do you find the exact value of \[arctan1\text{ or }{{\tan }^{-1}}1\]?

Answer 1

Hint:
tan\[x\] is the ratio of \[perpendicular\] to \[base\] in a right angled triangle and if this ratio is \[1\] that means the two sides are equal and using the property of an isosceles triangle their corresponding angles are also equal and that is \[45{}^\circ \] each by applying the angle sum property.

Complete step by step solution:

As we know that the \[tangent\] function for acute angles can be viewed as the ratio of the opposite to the adjacent side of the angle.
\[\Rightarrow \tan A=\dfrac{perpendicular}{base}=\dfrac{a}{b}\]
Let the given value \[{{\tan }^{-1}}1\] be \[x\]
\[\Rightarrow {{\tan }^{-1}}1=x\]
Now taking \[\tan \]both sides
\[\Rightarrow \tan \left( {{\tan }^{-1}}1 \right)=x\]
\[\Rightarrow 1=\tan x\]
If the ratio is 1, it means that the triangle is a right angle isosceles
Therefore, \[m\angle A=m\angle B--(1)\]
Now, using the angle sum property of the triangle
\[\Rightarrow \angle A+\angle B+\angle C=\angle 180{}^\circ --(2)\]
Since \[\angle C=90{}^\circ --(3)\]
From Equation \[(1),(2),(3)\]
\[\begin{align}
  & \Rightarrow \angle A+\angle B=90{}^\circ \\
 & \Rightarrow \angle A=\angle B=45{}^\circ \\
\end{align}\]
 And therefore the corresponding angle is \[45{}^\circ \] degrees of \[\dfrac{\pi }{4}rad\].

Hence, \[\arctan 1=\dfrac{\pi }{4}\]

Note:
To find the exact value we will use the basic foundations of trigonometry. Normally for \[45{}^\circ \] or \[\dfrac{\pi}{4} radian\] and in general form \[(2n+1)\dfrac{\pi }{4}\] where \[n\in Z\] the length other two sides are equal and the domain of \[arctan1\text{ or }{{\tan }^{-1}}1\] is \[\left[ -1,1 \right]\] and the range is \[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\].