`DeltaPQR` is right angled at Q .If `angleP=45^(@)` , then find the value of `("cosec"R+(1)/(3))`.

Options

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

Answer

To solve the problem, we need to find the value of \( \csc R + \frac{1}{3} \) in the right-angled triangle \( \Delta PQR \) where \( \angle P = 45^\circ \) and the triangle is right-angled at \( Q \).

### Step-by-Step Solution:

1. **Identify the angles in triangle \( PQR \)**:
- Since \( \Delta PQR \) is a right triangle with \( \angle Q = 90^\circ \) and \( \angle P = 45^\circ \), we can find \( \angle R \) using the triangle sum property.
- The sum of angles in a triangle is \( 180^\circ \).

\[ \angle P + \angle Q + \angle R = 180^\circ \]

Substituting the known values:

\[ 45^\circ + 90^\circ + \angle R = 180^\circ \]

\[ \angle R = 180^\circ - 135^\circ = 45^\circ \]

2. **Calculate \( \csc R \)**:
- We know that \( \csc R \) is the cosecant of angle \( R \).
- Since \( \angle R = 45^\circ \):

\[ \csc R = \csc 45^\circ \]

The cosecant of \( 45^\circ \) is given by:

\[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]

3. **Add \( \frac{1}{3} \) to \( \csc R \)**:
- Now we need to find \( \csc R + \frac{1}{3} \):

\[ \csc R + \frac{1}{3} = \sqrt{2} + \frac{1}{3} \]

4. **Combine the terms**:
- To combine \( \sqrt{2} \) and \( \frac{1}{3} \), we need a common denominator. The common denominator here is \( 3 \):

\[ \sqrt{2} + \frac{1}{3} = \frac{3\sqrt{2}}{3} + \frac{1}{3} = \frac{3\sqrt{2} + 1}{3} \]

### Final Answer:
Thus, the value of \( \csc R + \frac{1}{3} \) is:

\[ \frac{3\sqrt{2} + 1}{3} \]