If `U_(n) = sin n theta*sec^(n)theta, V_(n) = cosntheta*sec^(n)theta `for n= 0,1,2,… then `V_(n) -V_(n-1)+U_(n-1)tantheta` =

Options

A. 0
B. 1
C. `sintheta`
D. `costheta`

Answer

To solve the problem, we need to evaluate the expression \( V_n - V_{n-1} + U_{n-1} \tan \theta \) using the definitions of \( U_n \) and \( V_n \).

Given:
- \( U_n = \sin(n\theta) \sec^n(\theta) \)
- \( V_n = \cos(n\theta) \sec^n(\theta) \)

### Step 1: Write the expression

We start with the expression:
\[ V_n - V_{n-1} + U_{n-1} \tan \theta \]

### Step 2: Substitute \( V_n \) and \( V_{n-1} \)

Substituting the definitions:
\[ V_n = \cos(n\theta) \sec^n(\theta) \]
\[ V_{n-1} = \cos((n-1)\theta) \sec^{n-1}(\theta) \]
Thus, we have:
\[ V_n - V_{n-1} = \cos(n\theta) \sec^n(\theta) - \cos((n-1)\theta) \sec^{n-1}(\theta) \]

### Step 3: Substitute \( U_{n-1} \)

Next, we substitute \( U_{n-1} \):
\[ U_{n-1} = \sin((n-1)\theta) \sec^{n-1}(\theta) \]
And since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we have:
\[ U_{n-1} \tan \theta = \sin((n-1)\theta) \sec^{n-1}(\theta) \cdot \frac{\sin \theta}{\cos \theta} \]

### Step 4: Combine the terms

Now, we can combine all the terms:
\[ V_n - V_{n-1} + U_{n-1} \tan \theta = \left( \cos(n\theta) \sec^n(\theta) - \cos((n-1)\theta) \sec^{n-1}(\theta) \right) + \left( \sin((n-1)\theta) \sec^{n-1}(\theta) \cdot \frac{\sin \theta}{\cos \theta} \right) \]

### Step 5: Factor out \( \sec^{n-1}(\theta) \)

We can factor out \( \sec^{n-1}(\theta) \):
\[ = \sec^{n-1}(\theta) \left( \cos(n\theta) \sec(\theta) - \cos((n-1)\theta) + \sin((n-1)\theta) \frac{\sin \theta}{\cos \theta} \right) \]

### Step 6: Simplify the expression

Now, we can simplify the expression inside the parentheses:
\[ = \sec^{n-1}(\theta) \left( \cos(n\theta) \sec(\theta) - \cos((n-1)\theta) + \frac{\sin((n-1)\theta) \sin(\theta)}{\cos(\theta)} \right) \]

### Step 7: Use trigonometric identities

Using the identity \( \cos(a + b) = \cos a \cos b - \sin a \sin b \):
\[ \cos(n\theta) \sec(\theta) - \left( \cos((n-1)\theta) \cos(\theta) - \sin((n-1)\theta) \sin(\theta) \right) \]

### Step 8: Final simplification

After applying the identity and simplifying, we find that:
\[ = 0 \]

Thus, the final result is:
\[ V_n - V_{n-1} + U_{n-1} \tan \theta = 0 \]

### Conclusion

The answer is \( 0 \).