Least integral value of `3(tan^(2)theta + cot^(2)theta)-8(tantheta + cot theta) +10, theta in (0,pi/2)` can be

Answer

To find the least integral value of the expression \(3(\tan^2 \theta + \cot^2 \theta) - 8(\tan \theta + \cot \theta) + 10\) for \(\theta \in (0, \frac{\pi}{2})\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ E = 3(\tan^2 \theta + \cot^2 \theta) - 8(\tan \theta + \cot \theta) + 10 \]
Using the identity \(\tan \theta + \cot \theta = t\), we can express \(\tan^2 \theta + \cot^2 \theta\) in terms of \(t\):
\[ \tan^2 \theta + \cot^2 \theta = (\tan \theta + \cot \theta)^2 - 2\tan \theta \cot \theta = t^2 - 2 \]
Thus, we can rewrite \(E\) as:
\[ E = 3(t^2 - 2) - 8t + 10 \]
Simplifying this gives:
\[ E = 3t^2 - 6 - 8t + 10 = 3t^2 - 8t + 4 \]

### Step 2: Find the minimum value of the quadratic
The expression \(E = 3t^2 - 8t + 4\) is a quadratic function in \(t\). The minimum value of a quadratic \(ax^2 + bx + c\) occurs at \(t = -\frac{b}{2a}\):
\[ t = -\frac{-8}{2 \cdot 3} = \frac{8}{6} = \frac{4}{3} \]

### Step 3: Substitute \(t\) back into the expression
Now, we substitute \(t = \frac{4}{3}\) back into the expression for \(E\):
\[ E\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) + 4 \]
Calculating this gives:
\[ E\left(\frac{4}{3}\right) = 3 \cdot \frac{16}{9} - \frac{32}{3} + 4 \]
\[ = \frac{48}{9} - \frac{96}{9} + \frac{36}{9} = \frac{48 - 96 + 36}{9} = \frac{-12}{9} = -\frac{4}{3} \]

### Step 4: Find the least integral value
The least integral value greater than or equal to \(-\frac{4}{3}\) is \(-1\).

### Conclusion
Thus, the least integral value of the expression is:
\[ \boxed{-1} \]