Answer
$\dfrac{1}{\sec x+\tan x}+\dfrac{1}{\sec x-\tan x}=2\sec x$
Work Step by Step
$\dfrac{1}{\sec x+\tan x}+\dfrac{1}{\sec x-\tan x}=2\sec x$
Evaluate the sum of fractions on the left side:
$\dfrac{\sec x-\tan x+\sec x+\tan x}{\sec^{2}x-\tan^{2}x}=2\sec x$
Simplify the numerator:
$\dfrac{2\sec x}{\sec^{2}x-\tan^{2}x}=2\sec x$
Replace $\sec^{2}x$ by $1+\tan^{2}x$:
$\dfrac{2\sec x}{1+\tan^{2}x-\tan^{2}x}=2\sec x$
Since $1+\tan^{2}x-\tan^{2}x=1$, the identity is proved:
$2\sec x=2\sec x$
