In a `Delta ABC` right - angled at B, AB : AC = 1 : 2. Then the value of `(cot A + tan C)/(sin B + cos B)` is:

Options

A. `(2)/(sqrt3)`
B. `(sqrt3 + 1)/(2)`
C. `(2sqrt2 - sqrt2)/(2)`
D. `sqrt3-1`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the triangle properties
Given triangle ABC is right-angled at B, and the ratio of sides AB and AC is given as 1:2.

### Step 2: Assign lengths to the sides
Let AB = x and AC = 2x. Since AB is opposite angle C and AC is the hypotenuse, we can denote:
- AB = x
- AC = 2x

### Step 3: Use the Pythagorean theorem
Since triangle ABC is a right triangle, we can use the Pythagorean theorem to find the length of BC:
\[ BC^2 + AB^2 = AC^2 \]
Substituting the values:
\[ BC^2 + x^2 = (2x)^2 \]
\[ BC^2 + x^2 = 4x^2 \]
\[ BC^2 = 4x^2 - x^2 \]
\[ BC^2 = 3x^2 \]
Taking the square root:
\[ BC = \sqrt{3}x \]

### Step 4: Find angles A and C
Using the sine function for angle C:
\[ \sin C = \frac{AB}{AC} = \frac{x}{2x} = \frac{1}{2} \]
This means:
\[ C = 30^\circ \]

Using the angle sum property of triangles:
\[ A + B + C = 180^\circ \]
Since B = 90°:
\[ A + 90^\circ + 30^\circ = 180^\circ \]
Thus:
\[ A = 60^\circ \]

### Step 5: Calculate cot A and tan C
Now we can find cot A and tan C:
\[ \cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A} \]
For A = 60°:
\[ \cot 60^\circ = \frac{1}{\sqrt{3}} \]

For angle C = 30°:
\[ \tan C = \frac{\sin C}{\cos C} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \]

### Step 6: Combine cot A and tan C
Now we can calculate:
\[ \cot A + \tan C = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]

### Step 7: Calculate sin B and cos B
Since B = 90°:
\[ \sin B = 1 \quad \text{and} \quad \cos B = 0 \]
Thus:
\[ \sin B + \cos B = 1 + 0 = 1 \]

### Step 8: Final calculation
Now we can find the value of the expression:
\[ \frac{\cot A + \tan C}{\sin B + \cos B} = \frac{\frac{2}{\sqrt{3}}}{1} = \frac{2}{\sqrt{3}} \]

### Conclusion
The final answer is:
\[ \frac{2}{\sqrt{3}} \]