State whether \[f(x)=\tan x-x\] is increasing or decreasing in its domain.

Hint: For the given function to say it is increasing or decreasing we have to do the first derivative. After doing the first derivative if the value is > 0 that means the function is increasing or else decreasing.

Complete step-by-step solution -
Given function is \[f(x)=\tan x-x\]
By doing the first derivative if the value is,
\[{{f}^{1}}\left( x \right)={{\sec }^{2}}x-1\]. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . (1)
We know that \[\sec x\ge 1\]\[\forall x\in R\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
\[\forall x\in R\] is the domain.
By squaring on both sides of (2) we get,
\[{{\sec }^{2}}x\ge 1\]\[\forall x\in R\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
By substituting (a) in (1) we get
\[{{f}^{1}}\left( x \right)={{\sec }^{2}}x-1\]\[\ge 0\]
From the above we can conclude that
\[{{f}^{1}}\left( x \right)\]\[\ge 0\]\[\forall x\in R\]
Therefore \[f(x)\] is increasing in the domain.

Note: This is a direct problem which is solved by taking the first derivative and knowing the domain of the function. From (2) we can say that it is a direct property of \[\sec x\]. If \[{{f}^{1}}\left( x \right)\] is greater than zero we say the function is increasing.