Answer
The solution of the given question is $\cos \ x$.
Work Step by Step
$\frac{\left( \text{sec x + tan x} \right)\left( \text{secx}\ -\ \text{tan}\ \text{x} \right)}{\sec \ x}$
By using the formula $\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , now, the above expression can be further simplified as:
$\frac{\left( \text{sec x + tan x} \right)\left( \text{secx}\ -\ \text{tan}\ \text{x} \right)}{\sec \ x}=\frac{\text{se}{{\text{c}}^{2}}x\ -\ {{\tan }^{2}}\ x}{\sec \ x}\ \ \ $
Apply the reciprocal identity of trigonometry, which is $\text{sec }x=\frac{1}{\text{cos }x}$; then the above expression can be further simplified as:
$\begin{align}
& \frac{\left( \sec \text{ }x\text{ + }\tan \text{ }x \right)\left( \sec x\ -\ \tan \ x \right)}{\sec \ x}=\frac{{{\sec }^{2}}\ -\ {{\tan }^{2}}\ x}{\sec \ x} \\
& =\ \frac{1}{\sec \ x} \\
& =\cos \ x\
\end{align}$
Conjecture: Left side is equal to cos x.
Thus, the left side of the expression is equal to the right side, which is
$\frac{\left( \sec \text{ }x\text{ + }\tan \text{ }x \right)\left( \sec x\ -\ \tan \ x \right)}{\sec \ x}=\ \cos \ x$.
