If `I_(n)=int_(0)^(pi//4) tan^(n)x dx, (ngt1` is an integer ), then (a) `I_(n)+I_(n-2)=1/(n+1)` (b) `I_(n)+I_(n-2)=1/(n-1)` (c) `I_(2)+I_(4),I_(6),…….` are in H.P. (d) `1/(2(n+1))ltI_(n)lt1/(2(n-1))`

Options

A. `I_(n)+I_(n-2)=1/(n+1)`
B. `I_(n)+I_(n-2)=1/(n-1)`
C. `I_(2)+I_(4),I_(6),…….` are in H.P.
D. `1/(2(n+1))ltI_(n)lt1/(2(n-1))`

Explanation

To solve the problem, we need to evaluate the integral \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) for \( n > 1 \) and determine which of the given options is correct.

### Step-by-Step Solution:

1. **Start with the integral definition**:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

2. **Use the identity for \(\tan^n x\)**:
We can express \(\tan^n x\) in terms of \(\tan^{n-2} x\):
\[ \tan^n x = \tan^{n-2} x \cdot \tan^2 x \]
Thus, we can rewrite the integral as:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \tan^2 x \, dx \]

3. **Substitute \(\tan^2 x\)**:
We know that \(\tan^2 x = \sec^2 x - 1\). Therefore, we can split the integral:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx \]
This gives us:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \cdot \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \]

4. **Change of variable**:
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). The limits change from \( x = 0 \) to \( x = \frac{\pi}{4} \) which corresponds to \( t = 0 \) to \( t = 1 \):
\[ I_n = \int_0^1 t^{n-2} \, dt - I_{n-2} \]

5. **Evaluate the integral**:
The integral \( \int_0^1 t^{n-2} \, dt \) can be evaluated as:
\[ \int_0^1 t^{n-2} \, dt = \frac{1}{n-1} \]
Therefore, substituting back, we have:
\[ I_n = \frac{1}{n-1} - I_{n-2} \]

6. **Rearranging gives us the desired relation**:
Rearranging the above equation gives:
\[ I_n + I_{n-2} = \frac{1}{n-1} \]

### Conclusion:
Thus, the correct option is:
(b) \( I_n + I_{n-2} = \frac{1}{n-1} \)