QUESTION IMAGE
Question
review the following derivation of the tangent double - angle identity. the steps are not listed in the correct order. what is the correct order of the steps used to derive the identity? step 1: $\tan(2x)=\frac{\frac{2sin(x)}{cos(x)}}{1 - \frac{sin^{2}(x)}{cos^{2}(x)}}$ step 2: $\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$ step 3: $\tan(2x)=\frac{2sin(x)cos(x)}{cos^{2}(x)-sin^{2}(x)}$ 4, 3, 5, 1, 2 4, 5, 3, 1, 2 3, 4, 5, 1, 2 3, 4, 1, 5, 2
Step1: Recall tan formula
$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$, and $\sin(2x) = 2\sin(x)\cos(x)$, $\cos(2x)=\cos^{2}(x)-\sin^{2}(x)$, so $\tan(2x)=\frac{2\sin(x)\cos(x)}{\cos^{2}(x)-\sin^{2}(x)}$ (Step 3).
Step2: Divide by $\cos^{2}(x)$
Dividing both numerator and denominator of $\frac{2\sin(x)\cos(x)}{\cos^{2}(x)-\sin^{2}(x)}$ by $\cos^{2}(x)$ gives $\tan(2x)=\frac{\frac{2\sin(x)}{\cos(x)}}{1 - \frac{\sin^{2}(x)}{\cos^{2}(x)}}$ (Step 1).
Step3: Use tan identity
Since $\tan(x)=\frac{\sin(x)}{\cos(x)}$, we get $\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$ (Step 2). So the correct order is 4, 3, 5, 1, 2.
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4, 3, 5, 1, 2