Given:
[tex]\frac{\sec ^2x-\tan ^2x}{\sec x+\tan x}[/tex]
Sol:
[tex]\frac{\sec ^2x-\tan ^2x}{\sec x+\tan x}[/tex]
Use the formula:
[tex](a+b)(a-b)=a^2-b^2[/tex]
Then:
[tex]\begin{gathered} =\frac{\sec ^2x-\tan ^2x}{\sec x+\tan x} \\ =\frac{(\sec x-\tan x)(\sec x+\tan x)}{\sec x+\tan x} \\ =\sec x-\tan x \end{gathered}[/tex]
So missing line is:
[tex]=\frac{(\sec x-\tan x)(\sec x+\tan x)}{\sec x+\tan x}[/tex]
