If `sinA+cosA=(3)/(5)`, then find the value of `sinA-cosA`.

Options

A. `(sqrt(41))/(5)`
B. `(9)/(4)`
C. `(5)/(sqrt(41))`
D. `(7)/(sqrt(41))`

Answer

To solve the problem where \( \sin A + \cos A = \frac{3}{5} \) and we need to find the value of \( \sin A - \cos A \), we can follow these steps:

### Step 1: Square both sides of the equation
We start by squaring the equation:
\[ (\sin A + \cos A)^2 = \left(\frac{3}{5}\right)^2 \]
This gives us:
\[ \sin^2 A + 2\sin A \cos A + \cos^2 A = \frac{9}{25} \]

### Step 2: Use the Pythagorean identity
We know that \( \sin^2 A + \cos^2 A = 1 \). Therefore, we can substitute this into our equation:
\[ 1 + 2\sin A \cos A = \frac{9}{25} \]

### Step 3: Isolate the term with sine and cosine
Now, we isolate \( 2\sin A \cos A \):
\[ 2\sin A \cos A = \frac{9}{25} - 1 \]
Convert 1 to a fraction with a denominator of 25:
\[ 2\sin A \cos A = \frac{9}{25} - \frac{25}{25} = \frac{9 - 25}{25} = \frac{-16}{25} \]

### Step 4: Find \( \sin A - \cos A \)
Now we can use the identity:
\[ (\sin A - \cos A)^2 = \sin^2 A + \cos^2 A - 2\sin A \cos A \]
Substituting the known values:
\[ (\sin A - \cos A)^2 = 1 - 2\sin A \cos A \]
Substituting \( 2\sin A \cos A = \frac{-16}{25} \):
\[ (\sin A - \cos A)^2 = 1 - \left(\frac{-16}{25}\right) \]
Again, convert 1 to a fraction:
\[ (\sin A - \cos A)^2 = \frac{25}{25} + \frac{16}{25} = \frac{41}{25} \]

### Step 5: Take the square root
Now, we take the square root to find \( \sin A - \cos A \):
\[ \sin A - \cos A = \sqrt{\frac{41}{25}} = \frac{\sqrt{41}}{5} \]

Thus, the value of \( \sin A - \cos A \) is:
\[ \frac{\sqrt{41}}{5} \]