If `thetain[-(pi)/9,-(pi)/(36)]` such that `f(theta)=tan(theta+(5pi)/(18))+cos(theta+(5pi)/(18))-tan(theta+(7pi)/9)` have maximum and minimum values as `l_(1)` and `l_(2)` respectlvely, then

Options

A. `l_(1)=11/(sqrt3)`
B. `l_(2)=(-3+2sqrt2)/2`
C. `l_(2)=(2sqrt2+1)/(sqrt2)`
D. `l_(1)=11/(2sqrt3)`

Answer

To solve the problem, we need to find the maximum and minimum values of the function \( f(\theta) = \tan\left(\theta + \frac{5\pi}{18}\right) + \cos\left(\theta + \frac{5\pi}{18}\right) - \tan\left(\theta + \frac{7\pi}{9}\right) \) over the interval \( \theta \in \left[-\frac{\pi}{9}, -\frac{\pi}{36}\right] \).

### Step 1: Rewrite the function
We start by rewriting the function in a more manageable form. We can express \( f(\theta) \) as:
\[ f(\theta) = \tan\left(\theta + \frac{5\pi}{18}\right) + \cos\left(\theta + \frac{5\pi}{18}\right) - \tan\left(\theta + \frac{7\pi}{9}\right) \]

### Step 2: Change of variable
Let \( \alpha = \theta + \frac{7\pi}{9} \). Then, we can express \( \theta \) in terms of \( \alpha \):
\[ \theta = \alpha - \frac{7\pi}{9} \]
Now, we need to determine the new limits for \( \alpha \):
- When \( \theta = -\frac{\pi}{9} \):
\[ \alpha = -\frac{\pi}{9} + \frac{7\pi}{9} = \frac{6\pi}{9} = \frac{2\pi}{3} \]
- When \( \theta = -\frac{\pi}{36} \):
\[ \alpha = -\frac{\pi}{36} + \frac{7\pi}{9} = -\frac{\pi}{36} + \frac{28\pi}{36} = \frac{27\pi}{36} = \frac{3\pi}{4} \]

Thus, the new interval for \( \alpha \) is \( \left[\frac{2\pi}{3}, \frac{3\pi}{4}\right] \).

### Step 3: Rewrite the function in terms of \( \alpha \)
Now, substituting \( \theta \) in \( f(\theta) \):
\[ f(\alpha) = \tan\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) + \cos\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) - \tan(\alpha) \]
We simplify \( \tan\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) \) and \( \cos\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) \).

### Step 4: Find the derivative
To find the maximum and minimum values, we need to compute the derivative \( f'(\alpha) \) and set it to zero:
\[ f'(\alpha) = \sec^2\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) + \sin\left(\alpha - \frac{7\pi}{9} + \frac{5\pi}{18}\right) - \sec^2(\alpha) \]

### Step 5: Solve for critical points
Set \( f'(\alpha) = 0 \) and solve for \( \alpha \) within the interval \( \left[\frac{2\pi}{3}, \frac{3\pi}{4}\right] \).

### Step 6: Evaluate endpoints and critical points
Evaluate \( f(\alpha) \) at the endpoints \( \alpha = \frac{2\pi}{3} \) and \( \alpha = \frac{3\pi}{4} \), and at any critical points found in the previous step.

### Step 7: Determine maximum and minimum values
Compare the values obtained from the evaluations to determine the maximum value \( l_1 \) and the minimum value \( l_2 \).

### Final Result
The maximum and minimum values \( l_1 \) and \( l_2 \) will be the respective values of \( f(\alpha) \) at the critical points and endpoints.

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