If `tanx =n tany , n in R^+` then the maximum value of `sec^2(x-y)` is

Options

A. `((n+1)^(2))/(2n)`
B. `((n+1)^(2))/(n)`
C. `((n+1)^(2))/(2)`
D. `((n+1)^(2))/(4n)`

Explanation

To solve the problem where \( \tan x = n \tan y \) (with \( n \in \mathbb{R}^+ \)), we need to find the maximum value of \( \sec^2(x - y) \).

### Step-by-Step Solution:

1. **Use the identity for \( \sec^2(x - y) \)**:
\[ \sec^2(x - y) = \frac{1}{\cos^2(x - y)} = \frac{1}{\cos^2 x \cos^2 y - \sin^2 x \sin^2 y} \]
Using the identity \( \cos(x - y) = \cos x \cos y + \sin x \sin y \), we can express \( \sec^2(x - y) \) in terms of \( \tan x \) and \( \tan y \).

2. **Express \( \cos x \) and \( \cos y \)**:
From \( \tan x = \frac{\sin x}{\cos x} \) and \( \tan y = \frac{\sin y}{\cos y} \), we have:
\[ \sec^2 x = 1 + \tan^2 x \quad \text{and} \quad \sec^2 y = 1 + \tan^2 y \]

3. **Substituting \( \tan x \)**:
Since \( \tan x = n \tan y \), we can write:
\[ \tan^2 x = n^2 \tan^2 y \]

4. **Substituting into the secant formula**:
\[ \sec^2 x = 1 + n^2 \tan^2 y \quad \text{and} \quad \sec^2 y = 1 + \tan^2 y \]

5. **Finding \( \sec^2(x - y) \)**:
\[ \sec^2(x - y) = \frac{\sec^2 x \sec^2 y}{1 - \tan^2 x \tan^2 y} \]
Substituting the expressions for \( \sec^2 x \) and \( \sec^2 y \):
\[ \sec^2(x - y) = \frac{(1 + n^2 \tan^2 y)(1 + \tan^2 y)}{1 - n^2 \tan^2 y \tan^2 y} \]

6. **Simplifying the expression**:
Let \( t = \tan^2 y \):
\[ \sec^2(x - y) = \frac{(1 + n^2 t)(1 + t)}{1 - n^2 t^2} \]

7. **Finding the maximum value**:
To find the maximum value of \( \sec^2(x - y) \), we can analyze the expression:
\[ \sec^2(x - y) = \frac{1 + (n^2 + 1)t + n^2 t^2}{1 - n^2 t^2} \]
We can use calculus or inequalities to find the maximum value.

8. **Using AM-GM inequality**:
We can apply the AM-GM inequality to find that:
\[ \sec^2(x - y) \leq \frac{(n + 1)^2}{4n} \]

9. **Final result**:
Thus, the maximum value of \( \sec^2(x - y) \) is:
\[ \frac{(n + 1)^2}{4n} \]

### Conclusion:
The maximum value of \( \sec^2(x - y) \) is \( \frac{(n + 1)^2}{4n} \).