If `sintheta+costheta=sqrt(2)sin(90^(@)-theta)`, then `cottheta=?`

Options

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

Answer

To solve the equation \( \sin \theta + \cos \theta = \sqrt{2} \sin(90^\circ - \theta) \) and find the value of \( \cot \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the equation**:
\[ \sin \theta + \cos \theta = \sqrt{2} \sin(90^\circ - \theta) \]
We know that \( \sin(90^\circ - \theta) = \cos \theta \). Thus, we can rewrite the equation as:
\[ \sin \theta + \cos \theta = \sqrt{2} \cos \theta \]

2. **Rearranging the equation**:
Subtract \( \cos \theta \) from both sides:
\[ \sin \theta = \sqrt{2} \cos \theta - \cos \theta \]
This simplifies to:
\[ \sin \theta = (\sqrt{2} - 1) \cos \theta \]

3. **Dividing both sides by \( \cos \theta \)** (assuming \( \cos \theta \neq 0 \)):
\[ \frac{\sin \theta}{\cos \theta} = \sqrt{2} - 1 \]
This gives us:
\[ \tan \theta = \sqrt{2} - 1 \]

4. **Finding \( \cot \theta \)**:
We know that \( \cot \theta = \frac{1}{\tan \theta} \). Therefore:
\[ \cot \theta = \frac{1}{\sqrt{2} - 1} \]

5. **Rationalizing the denominator**:
To rationalize \( \cot \theta \), multiply the numerator and denominator by \( \sqrt{2} + 1 \):
\[ \cot \theta = \frac{1 \cdot (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]

### Final Answer:
Thus, the value of \( \cot \theta \) is:
\[ \cot \theta = \sqrt{2} + 1 \]