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**Hint:**The above given problem is a pretty straight forward question and is very easy to solve. To solve this given problem we require some basic knowledge of trigonometry and trigonometric equations and general values. We also need to know about the domain and range of the trigonometric functions as well as their inverse functions. The function \[\tan \left( {{\sec }^{-1}}\left( x \right) \right)\] will evaluate to any value between \[\left( -\infty ,\infty \right)\] .

**Complete step by step solution:**

The domain and ranges of the following trigonometric functions are as follows,

Function | Domain | Range |

\[f\left( x \right)=\sin x\] | \[\left( -\infty ,+\infty \right)\] | \[\left[ -1,1 \right]\] |

\[f\left( x \right)=\cos x\] | \[\left( -\infty ,+\infty \right)\] | \[\left[ -1,1 \right]\] |

\[f\left( x \right)=\tan x\] | – all real numbers except \[\left( \dfrac{\pi }{2}+n\pi \right)\] | \[\left( -\infty ,+\infty \right)\] |

\[f\left( x \right)=\text{cosec}x\] | all real numbers except \[\left( n\pi \right)\] | \[(-\infty ,-1]\bigcup{[1,+\infty )}\] |

\[f\left( x \right)=\sec x\] | all real numbers except \[\left( \dfrac{\pi }{2}+n\pi \right)\] | \[(-\infty ,-1]\bigcup{[1,+\infty )}\] |

\[f\left( x \right)=\cot x\] | - all real numbers except \[\left( n\pi \right)\] | \[\left( -\infty ,+\infty \right)\] |

\[f\left( x \right)={{\sin }^{-1}}x\] | \[\left[ -1,1 \right]\] | \[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\] |

\[f\left( x \right)={{\cos }^{-1}}x\] | \[\left[ -1,1 \right]\] | \[\left[ 0,\pi \right]\] |

\[f\left( x \right)={{\tan }^{-1}}x\] | \[\left( -\infty ,+\infty \right)\] | \[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\] |

\[f\left( x \right)=\text{cose}{{\text{c}}^{-1}}x\] | \[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\] | \[\left[ -\pi ,-\dfrac{\pi }{2} \right]\bigcup{\left[ 0,\dfrac{\pi }{2} \right]}\] |

\[f\left( x \right)={{\sec }^{-1}}x\] | \[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\] | \[\left[ 0,\dfrac{\pi }{2} \right]\bigcup{\left[ \pi ,\dfrac{3\pi }{2} \right]}\] |

\[f\left( x \right)={{\cot }^{-1}}x\] | \[\left( -\infty ,+\infty \right)\] | \[\left[ 0,\pi \right]\] |

For \[y={{\sec }^{-1}}x\] in the domain \[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\] we can write \[x=\sec y\], and the given problem transforms to \[\tan \left( y \right)\]

We now start off with the solution of the given problem, and we do,

In this problem, first of all let us assume a right angle triangle which has hypotenuse of length x and base of length 1 units. Thus if we consider the base angle as y degrees, we can write \[x=\sec y\] . Apart from this, we can also write,

\[\tan y=\dfrac{\sqrt{{{x}^{2}}-1}}{1}\] , because we can evaluate that the value of the length of the perpendicular of the right angled triangle is \[\sqrt{{{x}^{2}}-1}\] units. Now, from the above equation we can very easily write that the value of y will be equal to,

\[y={{\tan }^{-1}}\left( \sqrt{{{x}^{2}}-1} \right)\] . Now putting this value of $y$ in the given problem equation, we get,

\[\begin{align}

& \tan \left( y \right) \\

& \Rightarrow \tan \left( {{\tan }^{-1}}\left( \sqrt{{{x}^{2}}-1} \right) \right) \\

\end{align}\]

Now, form this above intermediate equation, we can clearly write, it’s value as,

\[\sqrt{{{x}^{2}}-1}\]

**Thus our answer to the problem is \[\sqrt{{{x}^{2}}-1}\].**

**Note:**In these types of questions we need to be very careful about the domain and range of the given trigonometric function. We first of all need to convert the given form in such a way that the equation becomes a function of an inverse function i.e. of the form \[\tan \left( {{\tan }^{-1}} \right)\] . The problem can also be done in another way, which is graphically by plotting it in a graph paper and analysing it.

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