To determine whether [tex]\(\frac{1 - \sin(2x)}{1 + \sin(2x)} = \tan^2\left(\frac{\pi}{4} - x\right)\)[/tex], let's analyze both expressions and simplify them to see if they are equivalent.
### Simplifying [tex]\(\frac{1 - \sin(2x)}{1 + \sin(2x)}\)[/tex]:
Given:
[tex]\[
\frac{1 - \sin(2x)}{1 + \sin(2x)}
\][/tex]
### Simplifying [tex]\(\tan^2\left(\frac{\pi}{4} - x\right)\)[/tex]:
1. Recall that [tex]\(\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan(x)}{1 + \tan\left(\frac{\pi}{4}\right)\tan(x)}\)[/tex].
2. Since [tex]\(\tan\left(\frac{\pi}{4}\right) = 1\)[/tex], substitute [tex]\(\tan\left(\frac{\pi}{4}\right) \)[/tex]:
[tex]\[
\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan(x)}{1 + \tan(x)}
\][/tex]
3. Now square both sides:
[tex]\[
\tan^2\left(\frac{\pi}{4} - x\right) = \left(\frac{1 - \tan(x)}{1 + \tan(x)}\right)^2
\][/tex]
### Checking Equivalent of Both Simplified Expressions
1. Our goal is to see if:
[tex]\[
\frac{1 - \sin(2x)}{1 + \sin(2x)} = \left(\frac{1 - \tan(x)}{1 + \tan(x)}\right)^2
\][/tex]
Given these steps and simplifications, compare them:
- The first simplified expression: [tex]\(\frac{1 - \sin(2x)}{1 + \sin(2x)} \)[/tex]
- The second simplified expression: [tex]\(\cot^2(x + \frac{\pi}{4})\)[/tex]
### Conclusion
Upon analyzing both simplified expressions and their computations showed no equality. Therefore, they are not equivalent:
[tex]\[
\frac{1 - \sin(2x)}{1 + \sin(2x)} \neq \tan^2\left(\frac{\pi}{4} - x\right)
\][/tex]
Thus, the initial equation does not hold true:
[tex]\[
\frac{1 - \sin(2x)}{1 + \sin(2x)} \neq \tan^2\left(\frac{\pi}{4} - x\right)
\][/tex]
So, the proposition is false.
