If `sin theta + cos theta = a` and `sec theta + "cosec" theta = b`, then which of the following relation is true ?

Options

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

Answer

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

1. **Given Equations:**
- \( \sin \theta + \cos \theta = a \) (Equation 1)
- \( \sec \theta + \csc \theta = b \) (Equation 2)

2. **Rewrite Equation 2:**
- Recall that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \).
- Therefore, we can rewrite Equation 2 as:
\[ \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = b \]

3. **Combine the Fractions:**
- To combine the fractions, we take the least common multiple (LCM):
\[ \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = b \]
- This simplifies to:
\[ \sin \theta + \cos \theta = b \sin \theta \cos \theta \]

4. **Substitute Equation 1 into the New Equation:**
- From Equation 1, we know that \( \sin \theta + \cos \theta = a \).
- Substitute this into the equation:
\[ a = b \sin \theta \cos \theta \]

5. **Square Equation 1:**
- Now, we square Equation 1:
\[ (\sin \theta + \cos \theta)^2 = a^2 \]
- Expanding this gives:
\[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = a^2 \]
- Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2 \sin \theta \cos \theta = a^2 \]

6. **Express \( \sin \theta \cos \theta \):**
- Rearranging gives:
\[ 2 \sin \theta \cos \theta = a^2 - 1 \]
- Thus:
\[ \sin \theta \cos \theta = \frac{a^2 - 1}{2} \]

7. **Substituting back into the equation for \( b \):**
- Substitute \( \sin \theta \cos \theta \) back into the equation \( a = b \sin \theta \cos \theta \):
\[ a = b \left(\frac{a^2 - 1}{2}\right) \]

8. **Rearranging to find the relation:**
- Multiply both sides by 2:
\[ 2a = b(a^2 - 1) \]
- Rearranging gives:
\[ 2a = ba^2 - b \]
- This can be rearranged to:
\[ ba^2 - 2a - b = 0 \]

9. **Final Relation:**
- The relation we derived is:
\[ 2a = b(a^2 - 1) \]