If `sinA=x-cosAandsecA=y- "cosec" A` , then the value of `y(x^(2)-1)` is equal to :

Options

A. 3x
B. 2x
C. 2xy
D. 0

Answer

To solve the problem, we start with the given equations:

1. \( \sin A = x - \cos A \)
2. \( \sec A = y - \csc A \)

We need to find the value of \( y(x^2 - 1) \).

### Step 1: Express \( x \) and \( y \) in terms of \( \sin A \) and \( \cos A \)

From the first equation, we can rearrange it to find \( x \):
\[ x = \sin A + \cos A \]

From the second equation, we know that \( \sec A = \frac{1}{\cos A} \) and \( \csc A = \frac{1}{\sin A} \). Rearranging gives us:
\[ y = \sec A + \csc A + \cos A \]
Substituting the definitions of secant and cosecant:
\[ y = \frac{1}{\cos A} + \frac{1}{\sin A} + \cos A \]

### Step 2: Calculate \( x^2 \)

Now we calculate \( x^2 \):
\[ x^2 = (\sin A + \cos A)^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A \]
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \):
\[ x^2 = 1 + 2\sin A \cos A \]

### Step 3: Calculate \( x^2 - 1 \)

Now, we find \( x^2 - 1 \):
\[ x^2 - 1 = 2\sin A \cos A \]

### Step 4: Substitute \( y \) and \( x^2 - 1 \)

Now we substitute \( y \) and \( x^2 - 1 \) into \( y(x^2 - 1) \):
\[ y(x^2 - 1) = \left( \frac{1}{\cos A} + \frac{1}{\sin A} + \cos A \right)(2\sin A \cos A) \]

### Step 5: Distribute \( 2\sin A \cos A \)

Distributing gives us:
\[ y(x^2 - 1) = 2\sin A \cos A \cdot \frac{1}{\cos A} + 2\sin A \cos A \cdot \frac{1}{\sin A} + 2\sin A \cos A \cdot \cos A \]
This simplifies to:
\[ y(x^2 - 1) = 2\sin A + 2\cos A + 2\sin A \cos^2 A \]

### Step 6: Factor out the common term

Factoring out \( 2 \):
\[ y(x^2 - 1) = 2(\sin A + \cos A + \sin A \cos^2 A) \]

### Step 7: Substitute \( x \) back into the equation

Since \( x = \sin A + \cos A \):
\[ y(x^2 - 1) = 2x \]

### Final Answer

Thus, the value of \( y(x^2 - 1) \) is:
\[ \boxed{2x} \]