If `3 sin theta + 4 cos theta=5`, then find the value of `4 sin theta-3 cos theta`.

Answer

To solve the equation \(3 \sin \theta + 4 \cos \theta = 5\) and find the value of \(4 \sin \theta - 3 \cos \theta\), we can follow these steps:

### Step 1: Define the expression
Let \(x = 4 \sin \theta - 3 \cos \theta\).

### Step 2: Square both expressions
Square both sides of the equation:
\[ x^2 = (4 \sin \theta - 3 \cos \theta)^2 \]
Using the identity \((a - b)^2 = a^2 - 2ab + b^2\), we expand:
\[ x^2 = 16 \sin^2 \theta + 9 \cos^2 \theta - 24 \sin \theta \cos \theta \]
This gives us our first equation:
\[ x^2 = 16 \sin^2 \theta + 9 \cos^2 \theta - 24 \sin \theta \cos \theta \quad \text{(Equation 1)} \]

### Step 3: Square the given equation
Now, square the given equation \(3 \sin \theta + 4 \cos \theta = 5\):
\[ (3 \sin \theta + 4 \cos \theta)^2 = 5^2 \]
Expanding this using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\[ 9 \sin^2 \theta + 16 \cos^2 \theta + 24 \sin \theta \cos \theta = 25 \quad \text{(Equation 2)} \]

### Step 4: Add both equations
Now, we will add Equation 1 and Equation 2:
\[ (16 \sin^2 \theta + 9 \cos^2 \theta - 24 \sin \theta \cos \theta) + (9 \sin^2 \theta + 16 \cos^2 \theta + 24 \sin \theta \cos \theta) = x^2 + 25 \]
This simplifies to:
\[ (16 \sin^2 \theta + 9 \sin^2 \theta) + (9 \cos^2 \theta + 16 \cos^2 \theta) = x^2 + 25 \]
Combining like terms:
\[ 25 \sin^2 \theta + 25 \cos^2 \theta = x^2 + 25 \]

### Step 5: Use the Pythagorean identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ 25(1) = x^2 + 25 \]
This simplifies to:
\[ 25 = x^2 + 25 \]

### Step 6: Solve for \(x^2\)
Subtract 25 from both sides:
\[ 0 = x^2 \]
Thus, we find:
\[ x = 0 \]

### Conclusion
Therefore, the value of \(4 \sin \theta - 3 \cos \theta\) is:
\[ \boxed{0} \]