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

Options

A. `5*cos^(-1)((3)/(5))`
B. `5, sin^(-1)((3)/(5))`
C. `7, sin^(-1)((4)/(5))`
D. None of these

Explanation

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

### Step 1: Rewrite the left-hand side
We start with the left-hand side:
\[ 3 \cos \theta + 4 \sin \theta \]
We can express this in the form \(R \sin(\theta + \alpha)\) using the identity:
\[ R \sin(\theta + \alpha) = R (\sin \theta \cos \alpha + \cos \theta \sin \alpha) \]
This means we need to find \(R\) and \(\alpha\) such that:
\[ R \cos \alpha = 3 \quad \text{and} \quad R \sin \alpha = 4 \]

### Step 2: Calculate \(R\)
To find \(R\), we use the Pythagorean theorem:
\[ R = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Find \(\alpha\)
Now, we can find \(\alpha\) using the relationships we established:
\[ \cos \alpha = \frac{3}{R} = \frac{3}{5} \quad \text{and} \quad \sin \alpha = \frac{4}{R} = \frac{4}{5} \]
Thus, we can find \(\alpha\) using:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{3/5} = \frac{4}{3} \]
From this, we can find \(\alpha\) using the inverse tangent:
\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Final Values
From the calculations:
- The value of \(A\) is \(5\).
- The value of \(\alpha\) can be expressed as:
\[ \alpha = \sin^{-1}\left(\frac{4}{5}\right) \quad \text{or} \quad \alpha = \tan^{-1}\left(\frac{4}{3}\right) \]

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