If `sintheta+costheta=a and (sintheta+costheta)/(sinthetacostheta)=b`, then

Options

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

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ \sin \theta + \cos \theta = a \]
\[ \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = b \]

2. **Square the First Equation**:
We square both sides of the first equation:
\[ (\sin \theta + \cos \theta)^2 = a^2 \]
Expanding the left side using the identity \((x + y)^2 = x^2 + y^2 + 2xy\):
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = a^2 \]

3. **Use the Pythagorean Identity**:
We know from trigonometric identities that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting this into our equation gives:
\[ 1 + 2\sin \theta \cos \theta = a^2 \]
Rearranging this, we find:
\[ 2\sin \theta \cos \theta = a^2 - 1 \]
Therefore:
\[ \sin \theta \cos \theta = \frac{a^2 - 1}{2} \]

4. **Substitute into the Second Equation**:
Now we substitute \(\sin \theta \cos \theta\) into the second equation:
\[ \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = b \]
Substituting \(\sin \theta + \cos \theta = a\) and \(\sin \theta \cos \theta = \frac{a^2 - 1}{2}\):
\[ \frac{a}{\frac{a^2 - 1}{2}} = b \]
Simplifying this gives:
\[ \frac{2a}{a^2 - 1} = b \]

5. **Final Expression**:
Thus, we have derived the relationship:
\[ b = \frac{2a}{a^2 - 1} \]