Question

Proving Identities

Verify the identity$\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$

Short answer

The expression$\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$

Step 2. Concept used.

To demonstrate that the preceding statement is an identity, divide it into two parts, LHS and RHS. Separate the two when you've done separating them and simplify them independently. If the results of a given statement are the same, it is called an identity.

Step 3. Calculation.

Now, simplify LHS of $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$:

$\begin{array}{l}\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=\frac{\sec x-\tan x+\sec x+\tan x}{(\sec x+\tan x)(\sec x-\tan x)}\\ (a+b)(a-b)=a^2-b^2\\ =\frac{2\sec x}{{\sec }^{2}x-{\tan }^{2}x}\\ ({\sec }^{2}x-{\tan }^{2}x=1)\\ =2\sec x\end{array}$

RHS of the equation is$2\sec x$

Both RHS and LHS are equal. Hence it is an identity