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**Hint:**In this question, we have a trigonometric function and to solve this we used the formula for factor; power of ‘a’ is two \[\left( 2 \right)\] minus power of ‘b’ is two\[\left( 2 \right)\]. The expression of power of ‘a’ is two \[\left( 2 \right)\] minus power of ‘b’ is two \[\left( 2 \right)\]is called the difference of squares. And this formula is expressed as below.

\[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]

This is the formula for the difference of squares.

**Complete step by step answer:**

Let’s come to the question, in the question the data is given as below.

\[{\tan ^2}x - 1 = 0\]

Then we used the difference of the square formula. The above equation is written as.

\[\left( {\tan x - 1} \right)\left( {\tan x + 1} \right) = 0\]

We know that, if the product of any number of terms is equal to zero then one of the terms must equal to zero. Then,

\[

\tan x - 1 = 0 \\

\tan x = 1 \\

\]

And,

\[

\tan x + 1 = 0 \\

\tan x = - 1 \\

\]

We know that for tangent function,

\[ \Rightarrow \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( {\dfrac{{5\pi }}{4}} \right) = 1\]

And,

\[\tan \left( {\dfrac{{3\pi }}{4}} \right) = \tan \left( {\dfrac{{7\pi }}{4}} \right) = - 1\]

Then we find the value of\[x\]for tangent function.

Thus,

\[ \Rightarrow \tan x = 1 = \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( {\dfrac{{5\pi }}{4}} \right)\]

Thus, \[x = \left( {\dfrac{\pi }{4}} \right),\left( {\dfrac{{5\pi }}{4}} \right)\] for \[0\]to\[2\pi \].

And,

\[ \Rightarrow \tan x = - 1 = \tan \left( {\dfrac{{3\pi }}{4}} \right) = \tan \left( {\dfrac{{7\pi }}{4}} \right)\]

Thus, \[x = \left( {\dfrac{{3\pi }}{4}} \right),\left( {\dfrac{{7\pi }}{4}} \right)\] for \[0\] to \[2\pi \].

**Therefore, the value of \[x\]are \[\left( {\dfrac{\pi }{4}} \right),\left( {\dfrac{{5\pi }}{4}} \right),\left( {\dfrac{{3\pi }}{4}} \right),\left( {\dfrac{{7\pi }}{4}} \right)\] for domain \[\left( {0 - 2\pi } \right)\].**

**Note:**

We know that, we find the factor of \[\left( {{a^2} - {b^2}} \right)\]form type.

First, we want to calculate the factor of the above formula.

Then,

\[ \Rightarrow {a^2} - {b^2}\]

In the above expression, we can add and subtract the\[ab\]. By adding and subtracting the\[ab\], there is no effect in the above expression.

Then,

The above expression is written as below.

\[{a^2} - {b^2} + ab - ab\]

This is written as below form.

\[ \Rightarrow {a^2} - ab + ab - {b^2}\]

Then we take the common, in the first two we take the\[a\], is common and the last two we take the\[b\], is common.

Then the above expression is written as below.

\[a\left( {a - b} \right) + b\left( {a - b} \right)\]

By solving the above expression, the result would be as below.

\[\therefore \left( {a - b} \right)\left( {a + b} \right)\]

Then, we prove that \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\].

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