If `(1+x)/(1 -x)=cos2theta +isin2theta`, prove that that `x=itantheta`.

Answer

To prove that \( x = i \tan \theta \) given the equation

\[ \frac{1+x}{1-x} = \cos 2\theta + i \sin 2\theta, \]

we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:

\[ \frac{1+x}{1-x} = \cos 2\theta + i \sin 2\theta. \]

### Step 2: Use the component and dividend method
According to the component and dividend method, we can express the left-hand side as:

\[ \frac{(1+x) - (1-x)}{(1+x) + (1-x)}. \]

### Step 3: Simplify the numerator and denominator
Now simplify the numerator and denominator:

- The numerator becomes:
\[ (1+x) - (1-x) = 1 + x - 1 + x = 2x. \]

- The denominator becomes:
\[ (1+x) + (1-x) = 1 + x + 1 - x = 2. \]

Thus, we have:

\[ \frac{2x}{2} = x. \]

### Step 4: Equate the simplified expression to the right-hand side
Now we equate this to the right-hand side:

\[ x = \cos 2\theta + i \sin 2\theta. \]

### Step 5: Use trigonometric identities
Using the identities for cosine and sine, we can express \( \cos 2\theta \) and \( \sin 2\theta \):

- Recall that:
\[ \cos 2\theta = 1 - 2 \sin^2 \theta, \]
\[ \sin 2\theta = 2 \sin \theta \cos \theta. \]

### Step 6: Substitute these identities into the equation
Substituting these into our equation gives:

\[ x = (1 - 2 \sin^2 \theta) + i (2 \sin \theta \cos \theta). \]

### Step 7: Factor out common terms
Notice that we can factor out \( i \):

\[ x = 1 - 2 \sin^2 \theta + i (2 \sin \theta \cos \theta). \]

### Step 8: Express in terms of tangent
Now, we need to express \( x \) in terms of \( i \tan \theta \):

Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, we can rewrite \( x \) as:

\[ x = i \frac{2 \sin \theta \cos \theta}{\cos^2 \theta} = i \tan \theta. \]

### Conclusion
Thus, we have proved that:

\[ x = i \tan \theta. \]