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# Prove the trigonometric identity $\sec\left(x\right)=\frac{\sin\left(2x\right)}{\sin\left(x\right)}-\frac{\cos\left(2x\right)}{\cos\left(x\right)}$

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## Step-by-step Solution

Problem to solve:

$\sec\left(x\right)=\frac{\sin\left(2x\right)}{\sin\left(x\right)}-\frac{\cos\left(2x\right)}{\cos\left(x\right)}$

Choose the solving method

1

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

$\sec\left(x\right)=\frac{2\sin\left(x\right)\cos\left(x\right)}{\sin\left(x\right)}+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
2

Simplify the fraction $\frac{2\sin\left(x\right)\cos\left(x\right)}{\sin\left(x\right)}$ by $\sin\left(x\right)$

$\sec\left(x\right)=2\cos\left(x\right)+\frac{-\cos\left(2x\right)}{\cos\left(x\right)}$
3

Combine all terms into a single fraction with $\cos\left(x\right)$ as common denominator

$\sec\left(x\right)=\frac{2\cos\left(x\right)\cos\left(x\right)-\cos\left(2x\right)}{\cos\left(x\right)}$
4

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

$\sec\left(x\right)=\frac{2\cos\left(x\right)^2-\cos\left(2x\right)}{\cos\left(x\right)}$
5

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

$\sec\left(x\right)=\frac{2\cos\left(x\right)^2-\left(2\cos\left(x\right)^2-1\right)}{\cos\left(x\right)}$
6

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

$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$
7

Applying the trigonometric identity: $\displaystyle\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$

$\sec\left(x\right)=\sec\left(x\right)$
8

Since both sides of the equality are equal, we have proven the identity

true

true
$\sec\left(x\right)=\frac{\sin\left(2x\right)}{\sin\left(x\right)}-\frac{\cos\left(2x\right)}{\cos\left(x\right)}$

### Main topic:

Trigonometric Identities

~ 0.37 s