If `a(tantheta+cottheta)=1,sintheta+costheta=b` with `0^(@)ltthetalt90^(@)` , then a relation between a and b is

Options

A. `b^(2)=2(a+1)`
B. `b^(2)=2(a-1)`
C. `2a=b^(2)-1`
D. `2a=b^(2)+1`

Answer

To find a relation between \( a \) and \( b \) given the equations \( a(\tan \theta + \cot \theta) = 1 \) and \( \sin \theta + \cos \theta = b \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the first equation:**
\[ a(\tan \theta + \cot \theta) = 1 \]
Recall that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Therefore:
\[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \]

2. **Combine the terms:**
\[ \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} \]

3. **Substitute back into the equation:**
\[ a \cdot \frac{1}{\sin \theta \cos \theta} = 1 \]
This implies:
\[ a = \sin \theta \cos \theta \]

4. **Now consider the second equation:**
\[ \sin \theta + \cos \theta = b \]
Square both sides:
\[ (\sin \theta + \cos \theta)^2 = b^2 \]
Expanding the left side:
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = b^2 \]
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2\sin \theta \cos \theta = b^2 \]

5. **Substituting \( a \) into the equation:**
Since \( a = \sin \theta \cos \theta \):
\[ 1 + 2a = b^2 \]

6. **Rearranging gives the final relation:**
\[ 2a = b^2 - 1 \]

### Final Relation:
Thus, the relation between \( a \) and \( b \) is:
\[ 2a = b^2 - 1 \]