Question

How do you evaluate $\cot (\arctan \left({\displaystyle \frac{3}{5}}\right))$?

Answer 1

Hint: We explain the function $arc\tan \left(x\right)$. We express the inverse function of tan in the form of $arc\tan \left(x\right)={\tan }^{-1}x$. We draw the graph of $arc\tan \left(x\right)$ and the line $x={\displaystyle \frac{3}{5}}$ to find the intersection point. Thereafter we take the cot ratio of that angle to find the solution.

Complete step by step answer:
The given expression is the inverse function of trigonometric ratio tan.
The arcus function represents the angle which on ratio tan gives the value.
So, $arc\tan \left(x\right)={\tan }^{-1}x$. If $arc\tan \left(x\right)=\alpha$ then we can say $\tan \alpha =x$.
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $2\pi$.
The general solution for that value where $\tan \alpha =x$ will be $n\pi +\alpha ,n\in \mathbb{Z}$.
But for $arc\tan \left(x\right)$, we won’t find the general solution. We use the principal value. For ratio tan we have $-{\displaystyle \frac{\pi }{2}}\le arc\tan \left(x\right)\le {\displaystyle \frac{\pi }{2}}$.
The graph of the function is

We now place the value of $x={\displaystyle \frac{3}{5}}$ in the function of $arc\tan \left(x\right)$.
Let the angle be $\theta$ for which $arc\tan \left({\displaystyle \frac{3}{5}}\right)=\theta$. This gives $\tan \theta ={\displaystyle \frac{3}{5}}$.
For this we take the line of $x={\displaystyle \frac{3}{5}}$ and see the intersection of the line with the graph $arc\tan \left(x\right)$.

Putting the value in the graph of $arc\tan \left(x\right)$, we get $\theta =30.96$. (approx.)
We get the value of y coordinates as ${30.96}^{\circ }$.
Now we take $\cot (\arctan \left({\displaystyle \frac{3}{5}}\right))=\cot \left({30.96}^{\circ }\right)={\displaystyle \frac{5}{3}}$.

Therefore, the value of $\cot (\arctan \left({\displaystyle \frac{3}{5}}\right))$ is ${\displaystyle \frac{5}{3}}$.

Note: We can also apply the trigonometric identity where $\cot \theta ={\displaystyle \frac{1}{\tan \theta }}$. Also, in the exact solution domain of $-{\displaystyle \frac{\pi }{2}}\le \theta \le {\displaystyle \frac{\pi }{2}}$, $\tan \left({\tan }^{-1}x\right)=x$.using those identities we get
$\cot (\arctan \left({\displaystyle \frac{3}{5}}\right))={\displaystyle \frac{1}{\tan \left({\tan }^{-1}\left({\displaystyle \frac{3}{5}}\right)\right)}}={\displaystyle \frac{1}{{\displaystyle \frac{3}{5}}}}={\displaystyle \frac{5}{3}}$.