If tan `A = sqrt(2)-1`, then the value of cosec A . Sec A is equal to

Options

A. `(1)/(sqrt(2))`
B. `(2)/(sqrt(2))`
C. `2 sqrt(2)`
D. `(sqrt(3))/(2)`

Answer

To solve the problem where \( \tan A = \sqrt{2} - 1 \) and we need to find the value of \( \csc A \cdot \sec A \), we can follow these steps:

### Step 1: Understand the relationship of tangent
Given that \( \tan A = \frac{\text{opposite}}{\text{adjacent}} \), we can represent the opposite side as \( \sqrt{2} - 1 \) and the adjacent side as \( 1 \).

### Step 2: Set up the triangle
Let:
- Opposite side (BC) = \( \sqrt{2} - 1 \)
- Adjacent side (AB) = \( 1 \)

### Step 3: Calculate the hypotenuse using Pythagorean theorem
Using the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values:
\[ AC^2 = 1^2 + (\sqrt{2} - 1)^2 \]
Calculating \( (\sqrt{2} - 1)^2 \):
\[ (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]
So,
\[ AC^2 = 1 + (3 - 2\sqrt{2}) = 4 - 2\sqrt{2} \]

### Step 4: Find \( \csc A \) and \( \sec A \)
We know:
- \( \csc A = \frac{AC}{BC} \)
- \( \sec A = \frac{AC}{AB} \)

Calculating \( \csc A \):
\[ \csc A = \frac{AC}{BC} = \frac{\sqrt{4 - 2\sqrt{2}}}{\sqrt{2} - 1} \]

Calculating \( \sec A \):
\[ \sec A = \frac{AC}{AB} = \frac{\sqrt{4 - 2\sqrt{2}}}{1} = \sqrt{4 - 2\sqrt{2}} \]

### Step 5: Calculate \( \csc A \cdot \sec A \)
Now we can find:
\[ \csc A \cdot \sec A = \frac{\sqrt{4 - 2\sqrt{2}}}{\sqrt{2} - 1} \cdot \sqrt{4 - 2\sqrt{2}} \]
This simplifies to:
\[ \csc A \cdot \sec A = \frac{(4 - 2\sqrt{2})}{\sqrt{2} - 1} \]

### Step 6: Simplify the expression
To simplify \( \frac{(4 - 2\sqrt{2})}{\sqrt{2} - 1} \):
Multiply numerator and denominator by \( \sqrt{2} + 1 \):
\[ \frac{(4 - 2\sqrt{2})(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(4\sqrt{2} + 4 - 2\cdot2 - 2\sqrt{2})}{2 - 1} \]
This simplifies to:
\[ \frac{(2\sqrt{2} + 4 - 4)}{1} = 2\sqrt{2} \]

### Final Answer
Thus, the value of \( \csc A \cdot \sec A \) is:
\[ \boxed{2\sqrt{2}} \]