If `y=acos^(2)x+2bsinxcosx+csin^(2)xandz=asin^(2)x-2bsinxcosx+c cos^(2)x`, then

Options

A. `y+z=a+c`
B. `y+z=a+b`
C. `y+a=x+b`
D. none of these

Answer

To solve the problem, we need to evaluate the expressions for \(y\) and \(z\) given the equations:

1. \(y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x\)
2. \(z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x\)

We will use the value of \(x = 90^\circ\) to simplify our calculations.

### Step 1: Substitute \(x = 90^\circ\) into \(y\)

\[ \cos 90^\circ = 0 \quad \text{and} \quad \sin 90^\circ = 1 \]

Substituting these values into the equation for \(y\):

\[ y = a \cos^2(90^\circ) + 2b \sin(90^\circ) \cos(90^\circ) + c \sin^2(90^\circ) \]

This simplifies to:

\[ y = a \cdot 0 + 2b \cdot 1 \cdot 0 + c \cdot 1^2 \]
\[ y = 0 + 0 + c = c \]

### Step 2: Substitute \(x = 90^\circ\) into \(z\)

Now we substitute \(x = 90^\circ\) into the equation for \(z\):

\[ z = a \sin^2(90^\circ) - 2b \sin(90^\circ) \cos(90^\circ) + c \cos^2(90^\circ) \]

This simplifies to:

\[ z = a \cdot 1^2 - 2b \cdot 1 \cdot 0 + c \cdot 0^2 \]
\[ z = a - 0 + 0 = a \]

### Step 3: Add the results for \(y\) and \(z\)

Now we add the results from Step 1 and Step 2:

\[ y + z = c + a \]

Thus, we have:

\[ y + z = a + c \]

### Conclusion

The final result is:

\[ y + z = a + c \]