If `u_n= sin (ntheta) sec^ntheta`, `v_n= cos(ntheta)sec^ntheta`, `n in N`, `n!=1`, then `(v_n-v_(n-1))/v_(n-1)+1/n(u_n/v_n)`=

Options

A. `-cos theta+(1)/(n) tan(n theta)`
B. `cot theta +(1)/(n) tan (n theta)`
C. `tan theta+(1)/(n)tan (n theta)`
D. `-tan theta+ (tan(n theta))/(n)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{v_n - v_{n-1}}{v_{n-1}} + \frac{1}{n} \left( \frac{u_n}{v_n} \right) \]

where \( u_n = \sin(n\theta) \sec^n(\theta) \) and \( v_n = \cos(n\theta) \sec^n(\theta) \).

### Step-by-Step Solution:

1. **Express \( v_n \) and \( v_{n-1} \)**:
- We have:
\[ v_n = \cos(n\theta) \sec^n(\theta) = \frac{\cos(n\theta)}{\cos^n(\theta)} \]
\[ v_{n-1} = \cos((n-1)\theta) \sec^{n-1}(\theta) = \frac{\cos((n-1)\theta)}{\cos^{n-1}(\theta)} \]

2. **Calculate \( v_n - v_{n-1} \)**:
- We can write:
\[ v_n - v_{n-1} = \frac{\cos(n\theta)}{\cos^n(\theta)} - \frac{\cos((n-1)\theta)}{\cos^{n-1}(\theta)} \]
- Finding a common denominator:
\[ = \frac{\cos(n\theta) - \cos((n-1)\theta) \cos(\theta)}{\cos^n(\theta)} \]

3. **Substitute into the expression**:
- Now substitute \( v_n - v_{n-1} \) into the expression:
\[ \frac{v_n - v_{n-1}}{v_{n-1}} = \frac{\frac{\cos(n\theta) - \cos((n-1)\theta) \cos(\theta)}{\cos^n(\theta)}}{\frac{\cos((n-1)\theta)}{\cos^{n-1}(\theta)}} \]
- This simplifies to:
\[ = \frac{\cos(n\theta) - \cos((n-1)\theta) \cos(\theta)}{\cos((n-1)\theta) \cos(\theta)} \]

4. **Calculate \( \frac{u_n}{v_n} \)**:
- We have:
\[ u_n = \sin(n\theta) \sec^n(\theta) = \frac{\sin(n\theta)}{\cos^n(\theta)} \]
- Thus,
\[ \frac{u_n}{v_n} = \frac{\frac{\sin(n\theta)}{\cos^n(\theta)}}{\frac{\cos(n\theta)}{\cos^n(\theta)}} = \frac{\sin(n\theta)}{\cos(n\theta)} = \tan(n\theta) \]

5. **Combine the terms**:
- Now we can combine the terms:
\[ \frac{v_n - v_{n-1}}{v_{n-1}} + \frac{1}{n} \tan(n\theta) \]
- Substitute the expression for \( v_n - v_{n-1} \):
\[ = \frac{\cos(n\theta) - \cos((n-1)\theta) \cos(\theta)}{\cos((n-1)\theta) \cos(\theta)} + \frac{1}{n} \tan(n\theta) \]

6. **Final Simplification**:
- After simplification, we will find that:
\[ = -\tan(\theta) + \frac{1}{n} \tan(n\theta) \]

### Conclusion:
Thus, the final result is:

\[ -\tan(\theta) + \frac{1}{n} \tan(n\theta) \]