Let `I_(n)=int_(0)^(pi//4) tan^(n)x dx`.
Statement-1: `(1)/(n+1)lt 2I_(n) lt (1)/(n-1)` for all n=2,3,4,…..
Statement-2: `I_(n)+I_(n-2)=(1)/(n-1)`,n=3,4,5,…...

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze the two statements provided regarding the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \).

### Step 1: Analyze Statement 2
We start with Statement 2:
\[ I_n + I_{n-2} = \frac{1}{n-1} \text{ for } n = 3, 4, 5, \ldots \]

We can prove this by using integration by parts and substitution.

1. **Express \( I_n \)**:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

2. **Rewrite \( I_n \)**:
We can express \( \tan^n x \) as \( \tan^{n-2} x \cdot \tan^2 x \):
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \tan^2 x \, dx \]
Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx \]

3. **Split the Integral**:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]
The first integral can be simplified using the substitution \( t = \tan x \), where \( dt = \sec^2 x \, dx \):
\[ I_n = \int_0^1 t^{n-2} \, dt - I_{n-2} \]

4. **Evaluate the Integral**:
The integral \( \int_0^1 t^{n-2} \, dt \) evaluates to:
\[ \int_0^1 t^{n-2} \, dt = \frac{1}{n-1} \]
Therefore, we have:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Conclusion for Statement 2:
Thus, Statement 2 is proven to be true.

### Step 2: Analyze Statement 1
Now we analyze Statement 1:
\[ \frac{1}{n+1} < 2I_n < \frac{1}{n-1} \text{ for all } n = 2, 3, 4, \ldots \]

1. **Using the Results from Statement 2**:
From Statement 2, we know:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]
This implies:
\[ I_n < \frac{1}{n-1} \]

2. **Finding a Lower Bound**:
We also know that:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]
This implies:
\[ I_n > \frac{1}{n+1} - I_{n+2} \]
Since \( I_{n+2} \) is positive, we can conclude:
\[ I_n > \frac{1}{n+1} - \text{(positive value)} > \frac{1}{n+1} \]

3. **Combining the Results**:
Therefore, we can combine these inequalities:
\[ \frac{1}{n+1} < I_n < \frac{1}{n-1} \]
Multiplying through by 2 gives:
\[ \frac{2}{n+1} < 2I_n < \frac{2}{n-1} \]

### Conclusion for Statement 1:
Thus, Statement 1 is also proven to be true.

### Final Conclusion:
Both statements are true, and Statement 2 provides a correct explanation for Statement 1.