If PQ and PR are equal sides of an isosceles right angled `DeltaPQR`, then find the value of `tan(angleP/(2))+cotangleQ`

Options

A. 2
B. `2sqrt2`
C. `2sqrt3`
D. `3sqrt2`

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\angle P}{2}\right) + \cot(\angle Q) \) in the isosceles right triangle \( \Delta PQR \) where \( PQ = PR \).

### Step-by-Step Solution:

1. **Identify the Angles**:
In an isosceles right triangle, the angles are:
- \( \angle P = 90^\circ \) (the right angle)
- \( \angle Q = \angle R = 45^\circ \) (the two equal angles)

2. **Calculate \( \frac{\angle P}{2} \)**:
\[ \frac{\angle P}{2} = \frac{90^\circ}{2} = 45^\circ \]

3. **Calculate \( \tan\left(\frac{\angle P}{2}\right) \)**:
\[ \tan\left(45^\circ\right) = 1 \]

4. **Calculate \( \cot(\angle Q) \)**:
Since \( \angle Q = 45^\circ \):
\[ \cot(45^\circ) = 1 \]

5. **Combine the Results**:
Now, we add the two results:
\[ \tan\left(\frac{\angle P}{2}\right) + \cot(\angle Q) = 1 + 1 = 2 \]

### Final Answer:
The value of \( \tan\left(\frac{\angle P}{2}\right) + \cot(\angle Q) \) is \( 2 \).

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