If `sin theta=(21)/(29)`, prove that `sectheta+tan theta=2(1)/(2)` if `theta` lies between 0 and `(pi)/(2)`. What will be the value of the expression when `theta` lies (i) between `(pi)/(2)` and `pi` and (ii) between `pi` and `(3pi)/(2)` ?

Answer

To solve the problem step by step, we start with the given information and apply trigonometric identities and properties.

### Step 1: Given Information
We know that:
\[ \sin \theta = \frac{21}{29} \]
This means that in a right triangle, the opposite side (perpendicular) is 21 and the hypotenuse is 29.

### Step 2: Find the Base
Using the Pythagorean theorem, we can find the adjacent side (base):
\[ \text{Hypotenuse}^2 = \text{Perpendicular}^2 + \text{Base}^2 \]
Substituting the known values:
\[ 29^2 = 21^2 + \text{Base}^2 \]
Calculating the squares:
\[ 841 = 441 + \text{Base}^2 \]
Rearranging gives:
\[ \text{Base}^2 = 841 - 441 = 400 \]
Taking the square root:
\[ \text{Base} = 20 \]

### Step 3: Calculate \( \sec \theta \) and \( \tan \theta \)
Now we can find \( \sec \theta \) and \( \tan \theta \):
\[ \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{29}{20} \]
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{21}{20} \]

### Step 4: Find \( \sec \theta + \tan \theta \)
Now we add \( \sec \theta \) and \( \tan \theta \):
\[ \sec \theta + \tan \theta = \frac{29}{20} + \frac{21}{20} = \frac{29 + 21}{20} = \frac{50}{20} = \frac{5}{2} \]
Thus, we have:
\[ \sec \theta + \tan \theta = 2 \cdot \frac{1}{2} \]
This proves the first part of the question.

### Step 5: Evaluate for \( \theta \) between \( \frac{\pi}{2} \) and \( \pi \)
In this range, we are in the second quadrant:
- Here, \( \sec \theta \) is negative, and \( \tan \theta \) is also negative.
\[ \sec \theta = -\frac{29}{20}, \quad \tan \theta = -\frac{21}{20} \]
Calculating \( \sec \theta + \tan \theta \):
\[ \sec \theta + \tan \theta = -\frac{29}{20} - \frac{21}{20} = -\frac{29 + 21}{20} = -\frac{50}{20} = -\frac{5}{2} \]

### Step 6: Evaluate for \( \theta \) between \( \pi \) and \( \frac{3\pi}{2} \)
In this range, we are in the third quadrant:
- Here, \( \sec \theta \) is negative, and \( \tan \theta \) is positive.
\[ \sec \theta = -\frac{29}{20}, \quad \tan \theta = \frac{21}{20} \]
Calculating \( \sec \theta + \tan \theta \):
\[ \sec \theta + \tan \theta = -\frac{29}{20} + \frac{21}{20} = \frac{-29 + 21}{20} = \frac{-8}{20} = -\frac{2}{5} \]

### Final Results
1. For \( \theta \) between \( 0 \) and \( \frac{\pi}{2} \):
\[ \sec \theta + \tan \theta = 2 \cdot \frac{1}{2} \]
2. For \( \theta \) between \( \frac{\pi}{2} \) and \( \pi \):
\[ \sec \theta + \tan \theta = -\frac{5}{2} \]
3. For \( \theta \) between \( \pi \) and \( \frac{3\pi}{2} \):
\[ \sec \theta + \tan \theta = -\frac{2}{5} \]