Question

How do you simplify the given term $\dfrac{{{\sec }^{2}}x-1}{\sec x-1}$?

Answer 1

We start solving the problem by equating the given division to a variable. We then make the necessary arrangements in that term and make use of the fact that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\times \left( a+b \right)$ to proceed through the problem. We then check whatever the common factors present in both numerator and denominator. We then cancel the common factors present in the numerator and denominator to get the required answer.

Complete step-by-step solution:
According to the problem, we are asked to find the simplified form of the given term $\dfrac{{{\sec }^{2}}x-1}{\sec x-1}$.
Let us assume $d=\dfrac{{{\sec }^{2}}x-1}{\sec x-1}$.
$\Rightarrow d=\dfrac{{{\sec }^{2}}x-{{1}^{2}}}{\sec x-1}$ ---(1).
We can see that the numerator in the equation (1) resembles the form $\left( {{a}^{2}}-{{b}^{2}} \right)$. We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\times \left( a+b \right)$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{\left( \sec x-1 \right)\times \left( \sec x+1 \right)}{\sec x-1}$ ---(2).
From the equation (2), we can see that the numerator and denominator have a common factor $\left( \sec x-1 \right)$. Let us cancel this factor from both numerator and denominator from equation (2) to proceed through the problem.
$\Rightarrow d=\sec x+1$.
We have found the simplified form of the given division $\dfrac{{{\sec }^{2}}x-1}{\sec x-1}$ as $\sec x+1$.
$\therefore$ The simplified form of the given division $\dfrac{{{\sec }^{2}}x-1}{\sec x-1}$ is $\sec x+1$.

Note: We can simplify the obtained result further as shown below:
We have found the result as $d=\sec x+1$ ---(3).
We know that $\sec \theta =\dfrac{1}{\cos \theta }$. Let us use this result in equation (3).
$\Rightarrow d=\dfrac{1}{\cos x}+1$.
$\Rightarrow d=\dfrac{1+\cos x}{\cos x}$ ---(4).
We know that $1+\cos x=2{{\cos }^{2}}\left( \dfrac{x}{2} \right)$. Let us use this result in equation (4).
$\Rightarrow d=\dfrac{2{{\cos }^{2}}\left( \dfrac{x}{2} \right)}{\cos x}$.
So, the simplified form is $\dfrac{2{{\cos }^{2}}\left( \dfrac{x}{2} \right)}{\cos x}$.