In ` Delta PQR` , if `angle PQR = 90^(@), PR = sqrt"" 5` and `PQ - RQ = 2` then what is the value of `sin R - cos R` ?

Options

A. ` (2)/( sqrt(5))`
B. ` ( sqrt(5))/(2)`
C. `0`
D. `1`

Answer

To solve the problem step by step, we will analyze the triangle PQR, where angle PQR is 90 degrees, PR is given as \( \sqrt{5} \), and the difference between sides PQ and RQ is 2.

### Step 1: Set up the triangle
We have triangle PQR with:
- Angle PQR = 90°
- PR (hypotenuse) = \( \sqrt{5} \)
- Let PQ = x and RQ = y.

### Step 2: Write the equation based on given information
From the problem, we know:
\[ PQ - RQ = 2 \]
This can be expressed as:
\[ x - y = 2 \]
From this, we can express \( x \) in terms of \( y \):
\[ x = y + 2 \]

### Step 3: Apply the Pythagorean theorem
Since angle PQR is 90 degrees, we can apply the Pythagorean theorem:
\[ PR^2 = PQ^2 + RQ^2 \]
Substituting the known values:
\[ (\sqrt{5})^2 = x^2 + y^2 \]
This simplifies to:
\[ 5 = x^2 + y^2 \]

### Step 4: Substitute \( x \) into the Pythagorean equation
Now substitute \( x = y + 2 \) into the equation:
\[ 5 = (y + 2)^2 + y^2 \]
Expanding this gives:
\[ 5 = (y^2 + 4y + 4) + y^2 \]
Combining like terms:
\[ 5 = 2y^2 + 4y + 4 \]

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ 2y^2 + 4y + 4 - 5 = 0 \]
This simplifies to:
\[ 2y^2 + 4y - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 4, c = -1 \):
\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ y = \frac{-4 \pm \sqrt{16 + 8}}{4} \]
\[ y = \frac{-4 \pm \sqrt{24}}{4} \]
\[ y = \frac{-4 \pm 2\sqrt{6}}{4} \]
\[ y = \frac{-2 \pm \sqrt{6}}{2} \]

### Step 7: Find values of \( y \) and \( x \)
Taking the positive root (since lengths cannot be negative):
\[ y = \frac{-2 + \sqrt{6}}{2} \]
Now substituting back to find \( x \):
\[ x = y + 2 = \frac{-2 + \sqrt{6}}{2} + 2 = \frac{-2 + \sqrt{6} + 4}{2} = \frac{2 + \sqrt{6}}{2} \]

### Step 8: Calculate \( \sin R \) and \( \cos R \)
Now we find \( \sin R \) and \( \cos R \):
- \( \sin R = \frac{PQ}{PR} = \frac{x}{\sqrt{5}} = \frac{\frac{2 + \sqrt{6}}{2}}{\sqrt{5}} = \frac{2 + \sqrt{6}}{2\sqrt{5}} \)
- \( \cos R = \frac{RQ}{PR} = \frac{y}{\sqrt{5}} = \frac{\frac{-2 + \sqrt{6}}{2}}{\sqrt{5}} = \frac{-2 + \sqrt{6}}{2\sqrt{5}} \)

### Step 9: Calculate \( \sin R - \cos R \)
Now we compute:
\[ \sin R - \cos R = \frac{2 + \sqrt{6}}{2\sqrt{5}} - \frac{-2 + \sqrt{6}}{2\sqrt{5}} \]
Combining the fractions:
\[ = \frac{(2 + \sqrt{6}) - (-2 + \sqrt{6})}{2\sqrt{5}} = \frac{2 + \sqrt{6} + 2 - \sqrt{6}}{2\sqrt{5}} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \]

### Final Answer
Thus, the value of \( \sin R - \cos R \) is:
\[ \frac{2}{\sqrt{5}} \]

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