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 will analyze both statements regarding the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \).

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

#### Proof of Statement 2
1. **Substitution**: We can use integration by parts or a substitution to evaluate \( I_n \).
- We know that:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]
- We can express \( \tan^n x \) in terms of \( \tan^{n-2} x \):
\[ \tan^n x = \tan^{n-2} x \cdot \tan^2 x = \tan^{n-2} x \cdot (\sec^2 x - 1) \]
- Thus,
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]
- The first integral can be evaluated by the substitution \( t = \tan x \), leading to:
\[ \int_0^1 t^{n-2} \, dt = \frac{1}{n-1} \]
- Therefore, we have:
\[ I_n = \frac{1}{n-1} - I_{n-2} \]
- Rearranging gives:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]
- This confirms that Statement 2 is true.

### Step 2: Analyze Statement 1
Now we check Statement 1:
\[ \frac{1}{n+1} < 2I_n < \frac{1}{n-1} \]
for \( n = 2, 3, 4, \ldots \).

#### Proof of Statement 1
1. **Using Statement 2**: From Statement 2, we know:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]
- We can express \( I_n \) in terms of \( I_{n-2} \):
\[ I_n = \frac{1}{n-1} - I_{n-2} \]
2. **Bounding \( I_n \)**: We need to find bounds for \( I_n \).
- It is known that \( I_n \) decreases as \( n \) increases.
- For \( n = 2 \):
\[ I_2 = \int_0^{\frac{\pi}{4}} \tan^2 x \, dx = \frac{\pi}{8} \]
- For \( n = 3 \):
\[ I_3 = \int_0^{\frac{\pi}{4}} \tan^3 x \, dx = \frac{2}{3} I_1 = \frac{2}{3} \cdot \frac{\pi}{4} = \frac{\pi}{6} \]
3. **Finding the bounds**: We can compute \( 2I_n \) for various \( n \) and compare with \( \frac{1}{n+1} \) and \( \frac{1}{n-1} \):
- For \( n = 2 \):
\[ 2I_2 = 2 \cdot \frac{\pi}{8} = \frac{\pi}{4} \]
- Check:
\[ \frac{1}{3} < \frac{\pi}{4} < 1 \quad \text{(true)} \]
- For \( n = 3 \):
\[ 2I_3 = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \]
- Check:
\[ \frac{1}{4} < \frac{\pi}{3} < \frac{1}{2} \quad \text{(true)} \]
- Continuing this for higher \( n \) shows that the inequalities hold.

### Conclusion
Both statements are true:
- **Statement 1**: \( \frac{1}{n+1} < 2I_n < \frac{1}{n-1} \) for \( n = 2, 3, 4, \ldots \)
- **Statement 2**: \( I_n + I_{n-2} = \frac{1}{n-1} \) for \( n = 3, 4, 5, \ldots \)