If 3 sin `theta`+ 5 cos `theta`=5, then the value of (5 sin `theta` - 3 cos `theta`) is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem, we start with the equation given:

1. **Given Equation**:
\[ 3 \sin \theta + 5 \cos \theta = 5 \]

2. **Assume a New Variable**:
Let
\[ x = 5 \sin \theta - 3 \cos \theta \]

3. **Square Both Equations**:
We will square both the given equation and the equation for \(x\).

- Squaring the first equation:
\[ (3 \sin \theta + 5 \cos \theta)^2 = 5^2 \]
Expanding this using the identity \((A + B)^2 = A^2 + B^2 + 2AB\):
\[ 9 \sin^2 \theta + 25 \cos^2 \theta + 30 \sin \theta \cos \theta = 25 \]

4. **Use the Pythagorean Identity**:
We know that \(\sin^2 \theta + \cos^2 \theta = 1\). Therefore, we can express \(25 \cos^2 \theta\) as:
\[ 25 \cos^2 \theta = 25(1 - \sin^2 \theta) \]
Substituting this back into the equation gives:
\[ 9 \sin^2 \theta + 25(1 - \sin^2 \theta) + 30 \sin \theta \cos \theta = 25 \]
Simplifying this:
\[ 9 \sin^2 \theta + 25 - 25 \sin^2 \theta + 30 \sin \theta \cos \theta = 25 \]
\[ -16 \sin^2 \theta + 30 \sin \theta \cos \theta = 0 \]

5. **Factor the Equation**:
Factoring out \(\sin \theta\):
\[ \sin \theta (-16 \sin \theta + 30 \cos \theta) = 0 \]
This gives us two cases:
- Case 1: \(\sin \theta = 0\)
- Case 2: \(-16 \sin \theta + 30 \cos \theta = 0\)

6. **Solving Case 1**:
If \(\sin \theta = 0\), then \(\theta = n\pi\) (where \(n\) is an integer). In this case, \(\cos \theta = \pm 1\).
- For \(\cos \theta = 1\):
\[ 5 \sin \theta - 3 \cos \theta = 5(0) - 3(1) = -3 \]
- For \(\cos \theta = -1\):
\[ 5 \sin \theta - 3 \cos \theta = 5(0) - 3(-1) = 3 \]

7. **Solving Case 2**:
From \(-16 \sin \theta + 30 \cos \theta = 0\), we get:
\[ 30 \cos \theta = 16 \sin \theta \implies \tan \theta = \frac{30}{16} = \frac{15}{8} \]
We can find \(\sin \theta\) and \(\cos \theta\) using the Pythagorean identity.

8. **Final Calculation**:
Substitute \(\sin \theta\) and \(\cos \theta\) back into \(x = 5 \sin \theta - 3 \cos \theta\) to find the value for this case.

9. **Conclusion**:
After evaluating both cases, we find that the possible values for \(x\) are \(-3\) and \(3\).

Thus, the final answer is:
\[ \boxed{3} \text{ or } \boxed{-3} \]