Start from the LHS (left-hand side)

Rewrite $\sec\left(x\right)$ in terms of sine and cosine

Start from the RHS (right-hand side)

Combine fractions with different denominator using the formula: $\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b\cdot c}{b\cdot d}$

Simplify $\sin\left(x\right)\cos\left(x\right)$ using the trigonometric identity: $\sin(2x)=2\sin(x)\cos(x)$

Divide fractions $\frac{\sin\left(2x\right)\cos\left(x\right)-\cos\left(2x\right)\sin\left(x\right)}{\frac{\sin\left(2x\right)}{2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

Multiply the single term $2$ by each term of the polynomial $\left(\sin\left(2x\right)\cos\left(x\right)-\cos\left(2x\right)\sin\left(x\right)\right)$

Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$

When multiplying two powers that have the same base ($\cos\left(x\right)$), you can add the exponents

Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$

Factor the polynomial $4\sin\left(x\right)\cos\left(x\right)^2-2\cos\left(2x\right)\sin\left(x\right)$ by it's greatest common factor (GCF): $2\sin\left(x\right)$

To prove an identity, we usually begin to work on the side of the equality that seems to be more complicated, or the side that is not expressed in terms of sine and cosine. In this problem, we will choose to work on the right side $\frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)}$ to reach the left side $\frac{1}{\cos\left(x\right)}$

Apply the trigonometric identity: $\cos\left(2\theta \right)$$=2\cos\left(\theta \right)^2-1$

Simplify the product $-(2\cos\left(x\right)^2-1)$

Multiply $-1$ times $-1$

Cancel like terms $2\cos\left(x\right)^2$ and $-2\cos\left(x\right)^2$

Since we have reached the expression of our goal, we have proven the identity