Question

How do you evaluate $\tan (arc\tan (10))$ ?

Answer 1

Hint:$\arctan$ is nothing but the inverse of $\tan$ . Firstly, we evaluate the trigonometric expression inside the brackets and then find the $\tan$ of that value. We can do this or else we can simply say that when $f({f^{ - 1}}(x)) = x$ Here in our case $\tan$ is $f(x)$ and $\arctan$ is ${f^{ - 1}}(x)$ .
Formula used:
Whenever there are a function and its inverse then, $f({f^{ - 1}}(x)) = x$

Complete step by step answer:The given trigonometric expression is, $\tan (arc\tan (10))$
We can also write the same expression as below.
$\Rightarrow \tan ({\tan ^{ - 1}}(10))$
Let us now consider $f(x) = \tan \theta$ and the inverse of the same function as, ${f^{ - 1}}(x) = {\tan ^{ - 1}}x$.
From the above question, we can say that $x$ is given the value of $10\;$ .
$\Rightarrow f({f^{ - 1}}(10)) = \tan ({\tan ^{ - 1}}(10))$
$\Rightarrow \tan ({\tan ^{ - 1}}(10)) = 10$
$\therefore$$\tan (\arctan (10))$ On evaluating we get the value as $10\;$.
Additional information: The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa.

Note:
We can also solve this by finding the actual values of performing the inverse function.
The given trigonometric expression is, $\tan (arc\tan (10))$
The value of $\arctan (10)$ or ${\tan ^{ - 1}}(10)$ is $84.2894\;$
We get this from the expression, $\tan x = 10$
We need to find the value of $x$ which is equal to ${\tan ^{ - 1}}(10)$ .
Now that we found the value of $\arctan (10)$ which is equal to $84.2894\;$ we put the value back in the expression.
$\Rightarrow \tan (\arctan (10)) = \tan (84.2894)$
Now again on finding the value of $\tan (84.2894)$ we get $10\;$
$\therefore \tan (\arctan (10)) = 10$
This also the proof for the formula we have used in the previous method which is, $f({f^{ - 1}}(x)) = x$