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Question: How do you evaluate \(\tan (arc\tan (10))\) ?...

How do you evaluate tan⁡(arctan⁡(10))\tan (arc\tan (10)) ?

Explanation

Solution

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

Complete step by step answer: The given trigonometric expression is, tan⁡(arctan⁡(10))\tan (arc\tan (10))
We can also write the same expression as below.
⇒tan⁡(tan⁡−1(10))\Rightarrow \tan ({\tan ^{ - 1}}(10))
Let us now consider f(x)=tan⁡θf(x) = \tan \theta and the inverse of the same function as, f−1(x)=tan⁡−1x{f^{ - 1}}(x) = {\tan ^{ - 1}}x.
From the above question, we can say that xx is given the value of 10  10\; .
⇒f(f−1(10))=tan⁡(tan⁡−1(10))\Rightarrow f({f^{ - 1}}(10)) = \tan ({\tan ^{ - 1}}(10))
⇒tan⁡(tan⁡−1(10))=10\Rightarrow \tan ({\tan ^{ - 1}}(10)) = 10
∴\therefore tan⁡(arctan⁡(10))\tan (\arctan (10)) On evaluating we get the value as 10  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⁡(arctan⁡(10))\tan (arc\tan (10))
The value of arctan⁡(10)\arctan (10) or tan⁡−1(10){\tan ^{ - 1}}(10) is 84.2894  84.2894\;
We get this from the expression, tan⁡x=10\tan x = 10
We need to find the value of xx which is equal to tan⁡−1(10){\tan ^{ - 1}}(10) .
Now that we found the value of arctan⁡(10)\arctan (10) which is equal to 84.2894  84.2894\; we put the value back in the expression.
⇒tan⁡(arctan⁡(10))=tan⁡(84.2894)\Rightarrow \tan (\arctan (10)) = \tan (84.2894)
Now again on finding the value of tan⁡(84.2894)\tan (84.2894) we get 10  10\;
∴tan⁡(arctan⁡(10))=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))=xf({f^{ - 1}}(x)) = x