If `3 cos theta+4 sin theta=A sin (theta+alpha)`, then values of A and `alpha ` are

Options

A. `V^(2)`
B. P
C. `PV^(2)`
D. RT

Explanation

To solve the equation \(3 \cos \theta + 4 \sin \theta = A \sin(\theta + \alpha)\), we will follow these steps:

### Step 1: Use the sine addition formula
We know that:
\[ \sin(\theta + \alpha) = \sin \theta \cos \alpha + \cos \theta \sin \alpha \]
Thus, we can rewrite the right-hand side of the equation:
\[ A \sin(\theta + \alpha) = A (\sin \theta \cos \alpha + \cos \theta \sin \alpha) = A \cos \alpha \sin \theta + A \sin \alpha \cos \theta \]

### Step 2: Rewrite the original equation
Now we can equate the coefficients of \(\sin \theta\) and \(\cos \theta\) from both sides of the equation:
\[ 3 \cos \theta + 4 \sin \theta = A \sin \alpha \cos \theta + A \cos \alpha \sin \theta \]

### Step 3: Compare coefficients
From the above equation, we can compare the coefficients:
- For \(\cos \theta\):
\[ 3 = A \sin \alpha \]
- For \(\sin \theta\):
\[ 4 = A \cos \alpha \]

### Step 4: Solve for \(A\)
We can express \(A\) in terms of \(\sin \alpha\) and \(\cos \alpha\):
1. From \(3 = A \sin \alpha\), we have:
\[ A = \frac{3}{\sin \alpha} \]
2. From \(4 = A \cos \alpha\), we have:
\[ A = \frac{4}{\cos \alpha} \]

### Step 5: Set the equations for \(A\) equal to each other
Equating the two expressions for \(A\):
\[ \frac{3}{\sin \alpha} = \frac{4}{\cos \alpha} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 3 \cos \alpha = 4 \sin \alpha \]
Dividing both sides by \(\cos \alpha\) (assuming \(\cos \alpha \neq 0\)):
\[ 3 = 4 \tan \alpha \]
Thus:
\[ \tan \alpha = \frac{3}{4} \]

### Step 7: Find \(\alpha\)
To find \(\alpha\), we can use the arctangent function:
\[ \alpha = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 8: Find \(A\)
Now, substituting \(\sin \alpha\) and \(\cos \alpha\) back to find \(A\):
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):
Let \(\sin \alpha = 3k\) and \(\cos \alpha = 4k\) for some \(k\):
\[ (3k)^2 + (4k)^2 = 1 \implies 9k^2 + 16k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5} \]
Thus:
\[ \sin \alpha = 3k = \frac{3}{5}, \quad \cos \alpha = 4k = \frac{4}{5} \]
Now substituting back to find \(A\):
\[ A = \frac{3}{\sin \alpha} = \frac{3}{\frac{3}{5}} = 5 \]

### Final Results
Thus, the values of \(A\) and \(\alpha\) are:
\[ A = 5, \quad \alpha = \tan^{-1}\left(\frac{3}{4}\right) \]