`DeltaPQR` is right angled at Q .If `mangleR=45^(@)`,then find the value of `("cosec"P-(sqrt(3))/(2))`.

Options

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

Answer

To solve the problem, we need to find the value of \( \csc P - \frac{\sqrt{3}}{2} \) given that triangle \( \Delta PQR \) is right-angled at \( Q \) and \( \angle R = 45^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles in the Triangle**:
Since \( \Delta PQR \) is a right-angled triangle at \( Q \), we know that:
\[ \angle P + \angle Q + \angle R = 180^\circ \]
Given \( \angle R = 45^\circ \) and \( \angle Q = 90^\circ \), we can find \( \angle P \):
\[ \angle P + 90^\circ + 45^\circ = 180^\circ \]
\[ \angle P + 135^\circ = 180^\circ \]
\[ \angle P = 180^\circ - 135^\circ = 45^\circ \]

2. **Calculate \( \csc P \)**:
We need to find \( \csc P \). Since we found that \( \angle P = 45^\circ \):
\[ \csc P = \csc 45^\circ \]
The cosecant of \( 45^\circ \) is:
\[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]

3. **Subtract \( \frac{\sqrt{3}}{2} \)**:
Now we need to calculate \( \csc P - \frac{\sqrt{3}}{2} \):
\[ \csc P - \frac{\sqrt{3}}{2} = \sqrt{2} - \frac{\sqrt{3}}{2} \]

4. **Simplify the Expression**:
To simplify \( \sqrt{2} - \frac{\sqrt{3}}{2} \), we can express \( \sqrt{2} \) with a common denominator:
\[ \sqrt{2} = \frac{2\sqrt{2}}{2} \]
Thus,
\[ \sqrt{2} - \frac{\sqrt{3}}{2} = \frac{2\sqrt{2} - \sqrt{3}}{2} \]

5. **Final Result**:
Therefore, the final value is:
\[ \csc P - \frac{\sqrt{3}}{2} = \frac{2\sqrt{2} - \sqrt{3}}{2} \]