In`DeltaABC,angleC=90^(@)andsecB=(5)/(4)`, then find all other trigonometric ratios for `angleB`.

Answer

To find all the trigonometric ratios for angle B in triangle ABC where angle C = 90° and sec B = 5/4, we can follow these steps:

### Step 1: Understand the given information
We know that:
- Angle C = 90°
- sec B = 5/4

### Step 2: Use the definition of secant
The secant of an angle is defined as the reciprocal of the cosine:
\[ \sec B = \frac{1}{\cos B} \]
Thus, we can find cos B:
\[ \cos B = \frac{4}{5} \]

### Step 3: Identify the sides of the triangle
In a right triangle, we can denote:
- Hypotenuse (AB) = 5 (since sec B = hypotenuse/base)
- Base (BC) = 4 (adjacent side to angle B)
- Perpendicular (AC) = ? (opposite side to angle B)

### Step 4: Use the Pythagorean theorem to find the perpendicular
According to the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]
Substituting the known values:
\[ 5^2 = AC^2 + 4^2 \]
\[ 25 = AC^2 + 16 \]
\[ AC^2 = 25 - 16 = 9 \]
\[ AC = \sqrt{9} = 3 \]

### Step 5: Calculate the remaining trigonometric ratios
Now that we have all three sides:
- Hypotenuse (AB) = 5
- Base (BC) = 4
- Perpendicular (AC) = 3

We can find the other trigonometric ratios:

1. **Sine (sin B)**:
\[ \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{3}{5} \]

2. **Tangent (tan B)**:
\[ \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} = \frac{3}{4} \]

3. **Cosecant (cosec B)**:
\[ \csc B = \frac{1}{\sin B} = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{5}{3} \]

4. **Secant (sec B)**:
\[ \sec B = \frac{1}{\cos B} = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{5}{4} \quad \text{(already given)} \]

5. **Cotangent (cot B)**:
\[ \cot B = \frac{1}{\tan B} = \frac{\text{Base}}{\text{Perpendicular}} = \frac{BC}{AC} = \frac{4}{3} \]

### Summary of Trigonometric Ratios for Angle B:
- \(\sin B = \frac{3}{5}\)
- \(\cos B = \frac{4}{5}\)
- \(\tan B = \frac{3}{4}\)
- \(\csc B = \frac{5}{3}\)
- \(\sec B = \frac{5}{4}\)
- \(\cot B = \frac{4}{3}\)