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

Last updated date: 22nd Feb 2024
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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 ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.