Let `tanx-tan^2x >0` and `|2sinx|<1` . Then the intersection of which of the following two sets satisfies both the inequalities?
`x > npi,n in Z` (b) `x > npi-pi/6,n in Z` `x

Options

A. `x gt npi, n in Z`
B. `x gt npi-pi//6, n in Z`
C. `x lt n pi-pi//4, n in Z`
D. `x lt npi+pi//6, n in Z`

Explanation

To solve the inequalities \( \tan x - \tan^2 x > 0 \) and \( |2\sin x| < 1 \), we will break down the problem step by step.

### Step 1: Solve the inequality \( \tan x - \tan^2 x > 0 \)

1. **Rearranging the inequality**:
\[ \tan x - \tan^2 x > 0 \]
This can be factored as:
\[ \tan x (1 - \tan x) > 0 \]

2. **Finding the critical points**:
The critical points occur when \( \tan x = 0 \) or \( \tan x = 1 \).
- \( \tan x = 0 \) gives \( x = n\pi \) for \( n \in \mathbb{Z} \)
- \( \tan x = 1 \) gives \( x = \frac{\pi}{4} + n\pi \) for \( n \in \mathbb{Z} \)

3. **Testing intervals**:
We will test the intervals determined by the critical points:
- Interval 1: \( (-\infty, n\pi) \)
- Interval 2: \( (n\pi, \frac{\pi}{4} + n\pi) \)
- Interval 3: \( (\frac{\pi}{4} + n\pi, (n+1)\pi) \)

We find that:
- In Interval 1: \( \tan x < 0 \) (not valid)
- In Interval 2: \( \tan x > 0 \) and \( 1 - \tan x > 0 \) (valid)
- In Interval 3: \( \tan x > 1 \) (not valid)

Thus, we conclude:
\[ 0 < \tan x < 1 \implies 0 < x < \frac{\pi}{4} + n\pi \]

### Step 2: Solve the inequality \( |2\sin x| < 1 \)

1. **Rearranging the inequality**:
\[ |2\sin x| < 1 \implies -1 < 2\sin x < 1 \]
Dividing by 2:
\[ -\frac{1}{2} < \sin x < \frac{1}{2} \]

2. **Finding the intervals**:
- The inequality \( \sin x < \frac{1}{2} \) gives:
\[ x < \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x > \frac{5\pi}{6} + 2n\pi \]
- The inequality \( \sin x > -\frac{1}{2} \) gives:
\[ x > -\frac{\pi}{6} + 2n\pi \quad \text{and} \quad x < \frac{7\pi}{6} + 2n\pi \]

Combining these, we have:
\[ -\frac{\pi}{6} + 2n\pi < x < \frac{\pi}{6} + 2n\pi \]

### Step 3: Finding the intersection of the two sets

1. **From the first inequality**:
\[ n\pi < x < n\pi + \frac{\pi}{4} \]

2. **From the second inequality**:
\[ -\frac{\pi}{6} + 2n\pi < x < \frac{\pi}{6} + 2n\pi \]

3. **Finding the intersection**:
We need to find \( x \) such that:
\[ n\pi < x < n\pi + \frac{\pi}{4} \]
and
\[ -\frac{\pi}{6} + 2n\pi < x < \frac{\pi}{6} + 2n\pi \]

The intersection will yield:
\[ n\pi + \frac{\pi}{6} < x < n\pi + \frac{\pi}{4} \]

### Conclusion

The valid intervals satisfying both inequalities are:
- \( n\pi < x < n\pi + \frac{\pi}{4} \)
- The options that satisfy these conditions are:
- \( x > n\pi - \frac{\pi}{6} \)
- \( x < n\pi + \frac{\pi}{6} \)

Thus, the correct options are:
- (a) \( x > n\pi \)
- (d) \( x < n\pi + \frac{\pi}{6} \)