Question

Find the value of $\dfrac{1}{\sec x-\tan x}-\dfrac{1}{\cos x}=?$

Answer 1

Hint:We will be using the concept of trigonometric function to solve the problem. We will be multiplying the term $\dfrac{1}{\sec x-\tan x}$ with $\sec x+\tan x$ in both numerator and denominator to simplify the expression then we will use the identity that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ to further simplify the question.

Complete step-by-step answer:

Now, we have to find the value of $\dfrac{1}{\sec x-\tan x}-\dfrac{1}{\cos x}=?$.
Now, we will first multiply the term $\dfrac{1}{\sec x-\tan x}$ with $\sec x+\tan x$ in both numerator and denominator. So, we have ${{\sec }^{2}}x+{{\tan }^{2}}x=1$ in the denominator of the first term.
So, we have,
$\dfrac{\sec x+\tan x}{\left( \sec x-\tan x \right)\left( \sec x+\tan x \right)}-\dfrac{1}{\cos x}$
Now, we know the trigonometric identity that,
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
So, we have on using this identity,
$\dfrac{\sec x+\tan x}{{{\sec }^{2}}x-{{\tan }^{2}}x}-\dfrac{1}{\cos x}$
Now, we know the identity that,
$\begin{align}
  & {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\
 & \dfrac{1}{\cos x}=\sec x \\
\end{align}$
So, on using this we have,
$\begin{align}
  & =\sec x+\tan x-\sec x \\
 & =\tan x \\
\end{align}$
So, we have the value of,
$\dfrac{1}{\sec x-\tan x}-\dfrac{1}{\cos x}=\tan x$.

Note: To solve these type of question it is important to note that we have first converted the term $\dfrac{1}{\sec x-\tan x}$ to $\sec x+\tan x$ by multiplying it with $\sec x+\tan x$ in both numerator and denominator and using the trigonometric identity that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$.Students should remember important trigonometric identities,reciprocals of trigonometric functions and formulas to solve these types of questions.