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To prove the trigonometric identity:
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
we start by simplifying both sides of the equation separately and then demonstrate their equivalence.
### Step 1: Simplify the Left-Hand Side
The left-hand side of the equation is:
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} \][/tex]
Recall that [tex]\(\cos(2\theta)\)[/tex] and [tex]\(\sin(2\theta)\)[/tex] can be expressed in terms of [tex]\(\tan(\theta)\)[/tex]:
[tex]\[ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \][/tex]
[tex]\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} \][/tex]
Substitute these into the left-hand side expression:
[tex]\[ \frac{\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}}{1 - \frac{2\tan(\theta)}{1 + \tan^2(\theta)}} \][/tex]
Combine the fractions:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \div \left( \frac{1 + \tan^2(\theta) - 2\tan(\theta)}{1 + \tan^2(\theta)} \right) \][/tex]
Simplify the division of fractions by multiplying by the reciprocal:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \times \frac{1 + \tan^2(\theta)}{1 + \tan^2(\theta) - 2\tan(\theta)} \][/tex]
The [tex]\(1 + \tan^2(\theta)\)[/tex] terms cancel out:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta) - 2\tan(\theta)} \][/tex]
This can be factored and rewritten as:
[tex]\[ \frac{-(\tan^2(\theta) - 1)}{-(2\tan(\theta) - 1 - \tan^2(\theta))} = \frac{-(1 - \tan^2(\theta))}{-(1 - \tan(\theta) + \tan(\theta))} = -\frac{1 - \tan^2(\theta)}{1 - 2\tan(\theta) + \tan^2(\theta)} \][/tex]
### Step 2: Simplify the Right-Hand Side
The right-hand side of the equation is:
[tex]\[ \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
### Step 3: Show Equivalence
By simplifying the left-hand side, we obtained:
[tex]\[ \frac{-\cos(2\theta)}{\sin(2\theta) - 1} \][/tex]
and the right-hand side is:
[tex]\[ -\frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
Since these forms are identical:
[tex]\[ \frac{-\cos(2\theta)}{\sin(2\theta) - 1} = -\frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
Thus, the simplified forms of both the left-hand side and right-hand side expressions match. Therefore, we have proven the given trigonometric identity.
Therefore,
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
we start by simplifying both sides of the equation separately and then demonstrate their equivalence.
### Step 1: Simplify the Left-Hand Side
The left-hand side of the equation is:
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} \][/tex]
Recall that [tex]\(\cos(2\theta)\)[/tex] and [tex]\(\sin(2\theta)\)[/tex] can be expressed in terms of [tex]\(\tan(\theta)\)[/tex]:
[tex]\[ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \][/tex]
[tex]\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} \][/tex]
Substitute these into the left-hand side expression:
[tex]\[ \frac{\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}}{1 - \frac{2\tan(\theta)}{1 + \tan^2(\theta)}} \][/tex]
Combine the fractions:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \div \left( \frac{1 + \tan^2(\theta) - 2\tan(\theta)}{1 + \tan^2(\theta)} \right) \][/tex]
Simplify the division of fractions by multiplying by the reciprocal:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \times \frac{1 + \tan^2(\theta)}{1 + \tan^2(\theta) - 2\tan(\theta)} \][/tex]
The [tex]\(1 + \tan^2(\theta)\)[/tex] terms cancel out:
[tex]\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta) - 2\tan(\theta)} \][/tex]
This can be factored and rewritten as:
[tex]\[ \frac{-(\tan^2(\theta) - 1)}{-(2\tan(\theta) - 1 - \tan^2(\theta))} = \frac{-(1 - \tan^2(\theta))}{-(1 - \tan(\theta) + \tan(\theta))} = -\frac{1 - \tan^2(\theta)}{1 - 2\tan(\theta) + \tan^2(\theta)} \][/tex]
### Step 2: Simplify the Right-Hand Side
The right-hand side of the equation is:
[tex]\[ \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
### Step 3: Show Equivalence
By simplifying the left-hand side, we obtained:
[tex]\[ \frac{-\cos(2\theta)}{\sin(2\theta) - 1} \][/tex]
and the right-hand side is:
[tex]\[ -\frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
Since these forms are identical:
[tex]\[ \frac{-\cos(2\theta)}{\sin(2\theta) - 1} = -\frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
Thus, the simplified forms of both the left-hand side and right-hand side expressions match. Therefore, we have proven the given trigonometric identity.
Therefore,
[tex]\[ \frac{\cos(2\theta)}{1 - \sin(2\theta)} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \][/tex]
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