How do you solve \[{\tan ^2}x - 1 = 0\].

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