If `z= costheta+isintheta` , then the value of `(z^(2n)-1)/(z^(2n)+1)` (A) `i tan n theta` (B) ` tan n theta` (C) `icot n theta` (D) `-i tan n theta`

Options

A. `icot n theta`
B. `I tan n theta`
C. `tan n theta`
D. `cot n theta`

Answer

To solve the problem, we start with the given expression for \( z \):

\[ z = \cos \theta + i \sin \theta \]

We need to find the value of:

\[ \frac{z^{2n} - 1}{z^{2n} + 1} \]

### Step 1: Express \( z^{2n} \)

Using Euler's formula, we can express \( z \) in exponential form:

\[ z = e^{i\theta} \]

Thus, we have:

\[ z^{2n} = (e^{i\theta})^{2n} = e^{i(2n\theta)} = \cos(2n\theta) + i \sin(2n\theta) \]

### Step 2: Substitute \( z^{2n} \) into the expression

Now substituting \( z^{2n} \) into the expression we need to evaluate:

\[ \frac{z^{2n} - 1}{z^{2n} + 1} = \frac{(\cos(2n\theta) + i \sin(2n\theta)) - 1}{(\cos(2n\theta) + i \sin(2n\theta)) + 1} \]

### Step 3: Simplify the numerator and denominator

The numerator simplifies to:

\[ \cos(2n\theta) - 1 + i \sin(2n\theta) \]

The denominator simplifies to:

\[ \cos(2n\theta) + 1 + i \sin(2n\theta) \]

### Step 4: Factor out the common terms

Now, we can factor out the common terms from the numerator and denominator. We can rewrite the numerator as:

\[ (\cos(2n\theta) - 1) + i \sin(2n\theta) \]

And the denominator as:

\[ (\cos(2n\theta) + 1) + i \sin(2n\theta) \]

### Step 5: Use trigonometric identities

Using the identity \( \cos(2n\theta) = 2\cos^2(n\theta) - 1 \) and \( \sin(2n\theta) = 2\sin(n\theta)\cos(n\theta) \), we can rewrite:

- \( \cos(2n\theta) - 1 = 2\cos^2(n\theta) - 2 = 2(\cos^2(n\theta) - 1) = -2\sin^2(n\theta) \)
- \( \cos(2n\theta) + 1 = 2\cos^2(n\theta) \)

### Step 6: Substitute back into the expression

Now substituting these back into our expression, we have:

\[ \frac{-2\sin^2(n\theta) + i(2\sin(n\theta)\cos(n\theta))}{2\cos^2(n\theta) + i(2\sin(n\theta)\cos(n\theta))} \]

### Step 7: Simplify further

Dividing both the numerator and denominator by 2 gives:

\[ \frac{-\sin^2(n\theta) + i\sin(n\theta)\cos(n\theta)}{\cos^2(n\theta) + i\sin(n\theta)\cos(n\theta)} \]

### Step 8: Multiply by the conjugate

To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(-\sin^2(n\theta) + i\sin(n\theta)\cos(n\theta))(\cos^2(n\theta) - i\sin(n\theta)\cos(n\theta))}{(\cos^2(n\theta) + i\sin(n\theta)\cos(n\theta))(\cos^2(n\theta) - i\sin(n\theta)\cos(n\theta))} \]

### Step 9: Final simplification

After simplification, we find that:

\[ \frac{i \tan(n\theta)}{1} = i \tan(n\theta) \]

Thus, the final result is:

\[ \frac{z^{2n} - 1}{z^{2n} + 1} = i \tan(n\theta) \]

### Conclusion

The correct answer is:

**(A) \( i \tan n \theta \)**