Question

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

Answer 1

Hint: 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.