Question

Proving Identities

Verify the identity${\sec }^{4}x-{\tan }^{4}x={\sec }^{2}x+{\tan }^{2}x$

Short answer

The expression${\sec }^{4}x-{\tan }^{4}x={\sec }^{2}x+{\tan }^{2}x$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression${\sec }^{4}x-{\tan }^{4}x={\sec }^{2}x+{\tan }^{2}x$

Step 2. Concept used.

To begin, divide the phrase into two sections: LHS and RHS. Separately, simplify both parts. If the results of both sections are the same, the expression is called an identity.

Step 3. Calculation.

Now, simplify LHS of ${\sec }^{4}x-{\tan }^{4}x={\sec }^{2}x+{\tan }^{2}x$:

$\begin{array}{l}\Rightarrow {\sec }^{4}x-{\tan }^{4}x\\ \because (a+b)(a-b)=a^2-b^2\\ \Rightarrow \left({\sec }^{2}x+{\tan }^{2}x\right)\left({\sec }^{2}x-{\tan }^{2}x\right)\\ \because \left({\sec }^{2}x-{\tan }^{2}x\right)=1\\ \Rightarrow {\sec }^{2}x+{\tan }^{2}x\end{array}$

RHS of the equation is${\sec }^{2}x+{\tan }^{2}x$

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