Let `tan^(-1) ( tan. (5pi)/(4))=alpha, tan^(-1) (-tan. (2pi)/(3))=beta`, then

Options

A. `alpha gt beta`
B. `4 alpha- 3 beta = 0`
C. `alpha + beta = ( 7 pi)/(12)`
D. None of these

Answer

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) and then establish the relationship between them.

### Step 1: Calculate \( \alpha \)

Given:
\[ \alpha = \tan^{-1}(\tan(5\pi/4)) \]

We can express \( 5\pi/4 \) as:
\[ 5\pi/4 = \pi + \pi/4 \]

Using the property of the tangent function, we know:
\[ \tan(\pi + x) = \tan(x) \]
Thus,
\[ \tan(5\pi/4) = \tan(\pi/4) \]

Now, since \( \tan(\pi/4) = 1 \), we have:
\[ \tan(5\pi/4) = 1 \]

Now substituting back into our equation for \( \alpha \):
\[ \alpha = \tan^{-1}(1) \]
\[ \alpha = \frac{\pi}{4} \]

### Step 2: Calculate \( \beta \)

Given:
\[ \beta = \tan^{-1}(-\tan(2\pi/3)) \]

We can express \( 2\pi/3 \) as:
\[ 2\pi/3 = \pi - \pi/3 \]

Using the property of the tangent function:
\[ \tan(\pi - x) = -\tan(x) \]
Thus,
\[ \tan(2\pi/3) = -\tan(\pi/3) \]

Since \( \tan(\pi/3) = \sqrt{3} \), we have:
\[ \tan(2\pi/3) = -\sqrt{3} \]

Now substituting back into our equation for \( \beta \):
\[ \beta = \tan^{-1}(-(-\sqrt{3})) = \tan^{-1}(\sqrt{3}) \]
\[ \beta = \frac{\pi}{3} \]

### Step 3: Establish the relationship between \( \alpha \) and \( \beta \)

From our calculations, we have:
\[ \alpha = \frac{\pi}{4} \]
\[ \beta = \frac{\pi}{3} \]

Now, we can find a relationship between \( \alpha \) and \( \beta \):
\[ \pi = 4\alpha \quad \text{and} \quad \pi = 3\beta \]

Setting these equal gives:
\[ 4\alpha = 3\beta \]

### Final Answer

Thus, the relationship between \( \alpha \) and \( \beta \) is:
\[ 4\alpha - 3\beta = 0 \]