Question

How do you simplify the expression $\left( {\sec x - 1} \right)\left( {\sec x + 1} \right)$ ?

Answer 1

In order to solve this question, first of all we will use one of the algebraic identity that is, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ . After that we will use one of the Pythagorean identity in the form of $\tan \theta$ and $\sec \theta$ that is, $1 + {\tan ^2}\theta = {\sec ^2}\theta$ and simplify it. Hence we will get the required result.

Complete step-by-step answer:
We have given the expression as
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right)$
And we are asked to simplify it.
So, first of all let the given expression as equation $\left( i \right)$
Therefore, we have
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right){\text{ }} - - - \left( i \right)$
Now we know that
$\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
Here, $a = \sec x$ and $b = 1$
Therefore, using the identity, we get
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right) = {\sec ^2}x - 1{\text{ }} - - - \left( {ii} \right)$
Now we know that the Pythagorean trigonometric identity in the form of $\tan \theta$ and $\sec \theta$ is given as: $1 + {\tan ^2}\theta = {\sec ^2}\theta$
On subtracting $1$ from both the sides, we get
$1 + {\tan ^2}\theta - 1 = {\sec ^2}\theta - 1$
On solving, we get
${\tan ^2}\theta = {\sec ^2}\theta - 1$
Now on comparing it with the equation $\left( {ii} \right)$ , we have
$\theta = x$
Therefore, we get the final result as
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right) = {\tan ^2}x$
Hence, the value of the expression $\left( {\sec x - 1} \right)\left( {\sec x + 1} \right)$ is equal to ${\tan ^2}x$

Note: While solving this question, if we don’t remember the formula of the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ then, we can also solve this expression by multiplying the terms and then applying the trigonometric identity.
Let’s solve it by this method:
We have given the expression as
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right)$
Now on multiplying the terms, we get
$= {\sec ^2}x + \sec x - \sec x - 1$
On solving, we get
$= {\sec ^2}x - 1$
Now we know that
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
$\Rightarrow {\tan ^2}\theta = {\sec ^2}\theta - 1$
Therefore, on comparing, we get the final result as
$\left( {\sec x - 1} \right)\left( {\sec x + 1} \right) = {\tan ^2}x$
Hence, we get the required answer.