If `sinA - sqrt(6)cos A= sqrt(7) cosA`, then the value of `cosA + sqrt(6)sinA` is:

Options

A. `sqrt(6)sinA`
B. `sqrt(7)sinA`
C. `sqrt(6)cosA`
D. `sqrt(7)cosA`

Answer

To solve the equation \( \sin A - \sqrt{6} \cos A = \sqrt{7} \cos A \) and find the value of \( \cos A + \sqrt{6} \sin A \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the initial equation:
\[ \sin A - \sqrt{6} \cos A - \sqrt{7} \cos A = 0 \]
Combine the cosine terms:
\[ \sin A - (\sqrt{6} + \sqrt{7}) \cos A = 0 \]

### Step 2: Isolate \(\sin A\)
From the rearranged equation, isolate \(\sin A\):
\[ \sin A = (\sqrt{6} + \sqrt{7}) \cos A \]

### Step 3: Use the Pythagorean Identity
We know from the Pythagorean identity that:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting \(\sin A\) from the previous step:
\[ ((\sqrt{6} + \sqrt{7}) \cos A)^2 + \cos^2 A = 1 \]
Expanding the equation:
\[ (\sqrt{6} + \sqrt{7})^2 \cos^2 A + \cos^2 A = 1 \]
\[ ((6 + 7) + 2\sqrt{42}) \cos^2 A + \cos^2 A = 1 \]
\[ (13 + 2\sqrt{42}) \cos^2 A = 1 \]

### Step 4: Solve for \(\cos^2 A\)
Now, solve for \(\cos^2 A\):
\[ \cos^2 A = \frac{1}{13 + 2\sqrt{42}} \]

### Step 5: Find \(\sin^2 A\)
Using the identity again:
\[ \sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{13 + 2\sqrt{42}} \]
This simplifies to:
\[ \sin^2 A = \frac{(13 + 2\sqrt{42}) - 1}{13 + 2\sqrt{42}} = \frac{12 + 2\sqrt{42}}{13 + 2\sqrt{42}} \]

### Step 6: Calculate \( \cos A + \sqrt{6} \sin A \)
Now we need to find \( \cos A + \sqrt{6} \sin A \):
\[ \cos A + \sqrt{6} \sin A = \cos A + \sqrt{6} \cdot (\sqrt{6} + \sqrt{7}) \cos A \]
This can be rewritten as:
\[ \cos A (1 + 6 + 6\sqrt{7}) = \cos A (7 + 6\sqrt{7}) \]

### Step 7: Substitute \(\cos A\)
Substituting \(\cos A\) from step 4:
\[ \cos A = \sqrt{\frac{1}{13 + 2\sqrt{42}}} \]
Thus:
\[ \cos A (7 + 6\sqrt{7}) = \sqrt{\frac{1}{13 + 2\sqrt{42}}} (7 + 6\sqrt{7}) \]

### Final Value
After simplifying, we find that:
\[ \cos A + \sqrt{6} \sin A = \sqrt{7} \]

### Conclusion
The value of \( \cos A + \sqrt{6} \sin A \) is \( \sqrt{7} \).