Sure, let's demonstrate the solution step-by-step to confirm that the equation [tex]\( 2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \)[/tex] is valid.
### Step 1: Understanding the Identity
We want to prove that:
[tex]\[ 2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
### Step 2: Recall Trigonometric Identities
First, let's recall some useful trigonometric identities:
1. [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]
2. [tex]\(\sin 2\theta = 2 \sin \theta \cos \theta\)[/tex]
### Step 3: Simplify the Left Side of the Equation
We know that:
[tex]\[ 2 \sin \theta \cos \theta \][/tex]
Using the double-angle formula for sine:
[tex]\[ 2 \sin \theta \cos \theta = \sin 2\theta \][/tex]
Therefore, the left side of the equation simplifies to:
[tex]\[ \sin 2\theta \][/tex]
### Step 4: Simplify the Right Side of the Equation
Now consider the right side:
[tex]\[ \frac{2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substitute this into the right side:
[tex]\[ \frac{2 \left( \frac{\sin \theta}{\cos \theta} \right)}{1 + \left( \frac{\sin \theta}{\cos \theta} \right)^2} \][/tex]
This simplifies to:
[tex]\[ \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} \][/tex]
Combine the fractions in the denominator:
[tex]\[ \frac{2 \frac{\sin \theta}{\cos \theta}}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} \][/tex]
Using the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ \frac{2 \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos^2 \theta}} \][/tex]
Simplify the denominator:
[tex]\[ 2 \frac{\sin \theta}{\cos \theta} \cdot \cos^2 \theta \][/tex]
[tex]\[ 2 \sin \theta \cdot \cos \theta \][/tex]
This simplifies to:
[tex]\[ 2 \sin \theta \cos \theta \][/tex]
Which, as we found earlier, is [tex]\(\sin 2\theta\)[/tex].
### Conclusion
Both the left side and the right side of the equation simplify to the same expression, [tex]\(\sin 2\theta\)[/tex]:
[tex]\[ 2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta \][/tex]
Therefore, the given equation is indeed valid. So, we have confirmed that:
[tex]\[ 2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
### Final Answer
Yes, the trigonometric identity is correct.
