Question

How do you find the exact value of $\cos \left( {{\tan }^{-1}}2 \right)$?

Answer 1

We explain the function $\arctan \left( x \right)$. We express the inverse function of tan in the form of $\arctan \left( x \right)={{\tan }^{-1}}x$. We draw the graph of $\arctan \left( x \right)$ and the line $x=2$ to find the intersection point. Thereafter we take the cos ratio of that angle to find the solution. We also use the representation of a right-angle triangle with height and base ratio being 2 and the angle being $\theta$.

Complete step by step solution:
The internal part ${{\tan }^{-1}}2$ of $\cos \left( {{\tan }^{-1}}2 \right)$ is an angle. We assume ${{\tan }^{-1}}2=\theta$.
This gives in ratio $\tan \theta =2$. We know $\tan \theta =\dfrac{\text{height}}{\text{base}}$.
We can take the representation of a right-angle triangle with height and base ratio being 2 and the angle being $\theta$. The height and base were considered with respect to that particular angle $\theta$.

In this case we take $AB=x$ and keeping the ratio in mind we have $AC=2x$ as the ratio has to be 8.
Now we apply the Pythagoras’ theorem to find the length of BC. $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$.
So, $B{{C}^{2}}={{x}^{2}}+{{\left( 2x \right)}^{2}}=5{{x}^{2}}$ which gives $BC=\sqrt{5}x$.
We need to find $\cos \left( {{\tan }^{-1}}2 \right)$ which is equal to $\cos \theta$.
This ratio gives $\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}$. So, $\cos \theta =\dfrac{AB}{BC}=\dfrac{x}{\sqrt{5}x}=\dfrac{1}{\sqrt{5}}$.
Therefore, $\cos \left( {{\tan }^{-1}}2 \right)$ is equal to $\dfrac{1}{\sqrt{5}}$.

Note: We can also apply the trigonometric image form to get the value of $\cos \left( {{\tan }^{-1}}2 \right)$.
It’s given that $\tan \theta =2$ and we need to find $\cos \theta$. We know $\cos \theta =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\theta }}$.
Putting the values, we get $\cos \theta =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\theta }}=\dfrac{1}{\sqrt{1+4}}=\dfrac{1}{\sqrt{5}}$.