In `/_\ABC`, right angled at A, `AC = 1.5 m, CB = 3 m` and `/_ABC=theta` find:
(i) `tan theta` (ii) `sectheta + "cosec"theta`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the sides of the triangle
In triangle ABC, we have:
- AC (perpendicular) = 1.5 m
- CB (base) = 3 m

### Step 2: Calculate the hypotenuse (AB)
Using the Pythagorean theorem:
\[ AB = \sqrt{AC^2 + CB^2} = \sqrt{(1.5)^2 + (3)^2} = \sqrt{2.25 + 9} = \sqrt{11.25} = 3.35 \, m \, (\text{approximately}) \]

### Step 3: Calculate \( \tan \theta \)
The tangent of angle \( \theta \) is given by:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{CB} = \frac{1.5}{3} = \frac{1}{2} \]

### Step 4: Calculate \( \sec \theta \) and \( \csc \theta \)
1. **Calculate \( \sec \theta \)**:
\[ \sec \theta = \frac{1}{\cos \theta} \]
To find \( \cos \theta \), we use:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CB}{AB} = \frac{3}{3.35} \approx 0.895 \]
Therefore,
\[ \sec \theta \approx \frac{1}{0.895} \approx 1.12 \]

2. **Calculate \( \csc \theta \)**:
\[ \csc \theta = \frac{1}{\sin \theta} \]
To find \( \sin \theta \), we use:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{1.5}{3.35} \approx 0.447 \]
Therefore,
\[ \csc \theta \approx \frac{1}{0.447} \approx 2.24 \]

### Step 5: Calculate \( \sec \theta + \csc \theta \)
\[ \sec \theta + \csc \theta \approx 1.12 + 2.24 \approx 3.36 \]

### Final Answers:
(i) \( \tan \theta = \frac{1}{2} \)
(ii) \( \sec \theta + \csc \theta \approx 3.36 \)

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