`I_(n)=int_(0)^(pi//4) tan^(n)x dx`, where `n in N`
Statement-1: `int_(0)^(pi//4) tan^(4)x dx=(3pi-8)/(12)`
Statement-2: `I_(n)+I_(n-2)=(1)/(n-1)`

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 evaluate the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) and verify the statements provided.

### Step 1: Express \( I_n \) in terms of \( I_{n-2} \)

We start with the integral:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

Using the identity \( \tan^n x = \tan^{n-2} x \cdot \tan^2 x \) and the fact that \( \tan^2 x = \sec^2 x - 1 \), we can rewrite the integral:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx \]
This expands to:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]
Thus, we have:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - I_{n-2} \]

### Step 2: Change of Variables

Now, we perform a change of variables. Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

This gives us:
\[ I_n = \int_0^1 t^{n-2} \, dt - I_{n-2} \]

### Step 3: Evaluate the Integral

Now we evaluate the integral:
\[ \int_0^1 t^{n-2} \, dt = \left[ \frac{t^{n-1}}{n-1} \right]_0^1 = \frac{1}{n-1} \]
Thus, we have:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Step 4: Verify Statement 2

This confirms Statement 2:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Step 5: Calculate \( I_4 \)

Now we need to calculate \( I_4 \) using Statement 2:
1. For \( n = 4 \):
\[ I_4 + I_2 = \frac{1}{3} \]
2. For \( n = 2 \):
\[ I_2 + I_0 = 1 \]
where \( I_0 = \int_0^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} \). Thus:
\[ I_2 + \frac{\pi}{4} = 1 \implies I_2 = 1 - \frac{\pi}{4} \]

### Step 6: Substitute \( I_2 \) into \( I_4 \)

Substituting \( I_2 \) back into the equation for \( I_4 \):
\[ I_4 + \left(1 - \frac{\pi}{4}\right) = \frac{1}{3} \]
This simplifies to:
\[ I_4 = \frac{1}{3} - 1 + \frac{\pi}{4} = \frac{\pi}{4} - \frac{2}{3} \]

### Step 7: Simplify \( I_4 \)

To combine the fractions:
\[ I_4 = \frac{\pi}{4} - \frac{2}{3} = \frac{3\pi - 8}{12} \]

### Conclusion

Thus, we have verified both statements:
- Statement 1: \( I_4 = \frac{3\pi - 8}{12} \) is true.
- Statement 2: \( I_n + I_{n-2} = \frac{1}{n-1} \) is true.