Certainly! Let's prove the given equation step-by-step.
We need to show that:
[tex]\[ \frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \tan x \][/tex]
### Step 1: Rewrite [tex]\(\sin 2x\)[/tex]
We know that:
[tex]\[ \sin 2x = 2 \sin x \cos x \][/tex]
### Step 2: Substitute [tex]\(\sin 2x\)[/tex]
Substitute [tex]\(\sin 2x\)[/tex] into the left-hand side of the original equation:
[tex]\[ \frac{2 \tan x - 2 \sin x \cos x}{2 \sin^2 x} \][/tex]
### Step 3: Simplify the numerator
Factor out the common factor of 2 in the numerator:
[tex]\[ \frac{2 (\tan x - \sin x \cos x)}{2 \sin^2 x} \][/tex]
Cancel the 2 on the numerator and the denominator:
[tex]\[ \frac{\tan x - \sin x \cos x}{\sin^2 x} \][/tex]
### Step 4: Rewrite [tex]\(\tan x\)[/tex]
We know that:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
Substituting [tex]\(\tan x\)[/tex] into the left-hand side:
[tex]\[ \frac{\frac{\sin x}{\cos x} - \sin x \cos x}{\sin^2 x} \][/tex]
### Step 5: Combine into a single fraction
Combine the terms in the numerator over a common denominator:
[tex]\[ \frac{\frac{\sin x - \sin x \cos^2 x}{\cos x}}{\sin^2 x} \][/tex]
Simplify the numerator:
[tex]\[ \frac{\sin x (1 - \cos^2 x)}{\cos x \sin^2 x} \][/tex]
### Step 6: Use the Pythagorean identity
We know that:
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
Substitute [tex]\(1 - \cos^2 x\)[/tex] with [tex]\(\sin^2 x\)[/tex]:
[tex]\[ \frac{\sin x \cdot \sin^2 x}{\cos x \sin^2 x} \][/tex]
### Step 7: Cancel [tex]\(\sin^2 x\)[/tex]
Cancel the [tex]\(\sin^2 x\)[/tex] term in the numerator and denominator:
[tex]\[ \frac{\sin x}{\cos x} \][/tex]
### Step 8: Simplify the fraction
Finally:
[tex]\[ \frac{\sin x}{\cos x} = \tan x \][/tex]
### Conclusion
We have shown that:
[tex]\[ \frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \tan x \][/tex]
Thus, we have proven the original equation as required.
