`x=sqrt(a^2cos^2alpha+b^2sin^2alpha)+sqrt(a^2sin^2alpha+b^2cos^2alpha)` then `x^2=a^2+b^2+2sqrt(p(a^2+b^2)-p^2),` where p can be is equal to

Options

A. `alpha^(2)cos^(2)alpha+b^(2)sin^(2)alpha`
B. `alpha^(2)sin^(2)alpha+b^(2) cos^(2)alpha`
C. `(1)/(2) [alpha^(2)+b^(2)+(alpha^(2)-b^(2))cos2alpha[`
D. `(1)/(2)[a^(2)+b^(2)-(a^(2)-b^(2))cos2alpha]`

Explanation

To solve the problem, we start with the expression given for \( x \):

\[ x = \sqrt{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha} + \sqrt{a^2 \sin^2 \alpha + b^2 \cos^2 \alpha} \]

### Step 1: Square the expression for \( x \)

We will square both sides to find \( x^2 \):

\[ x^2 = \left( \sqrt{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha} + \sqrt{a^2 \sin^2 \alpha + b^2 \cos^2 \alpha} \right)^2 \]

Using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \), we can expand this:

\[ x^2 = \left( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \right) + \left( a^2 \sin^2 \alpha + b^2 \cos^2 \alpha \right) + 2 \sqrt{(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)(a^2 \sin^2 \alpha + b^2 \cos^2 \alpha)} \]

### Step 2: Simplify the expression

Now, we can simplify the first two terms:

\[ x^2 = a^2 (\cos^2 \alpha + \sin^2 \alpha) + b^2 (\sin^2 \alpha + \cos^2 \alpha) \]

Using the Pythagorean identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):

\[ x^2 = a^2 + b^2 + 2 \sqrt{(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)(a^2 \sin^2 \alpha + b^2 \cos^2 \alpha)} \]

### Step 3: Define \( p \)

Let \( p = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \). Then we can express the square root term as:

\[ \sqrt{(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)(a^2 \sin^2 \alpha + b^2 \cos^2 \alpha)} = \sqrt{p(a^2 + b^2) - p^2} \]

### Step 4: Substitute back into the equation for \( x^2 \)

Now we can substitute this back into our expression for \( x^2 \):

\[ x^2 = a^2 + b^2 + 2\sqrt{p(a^2 + b^2) - p^2} \]

### Conclusion

Thus, we have shown that:

\[ x^2 = a^2 + b^2 + 2\sqrt{p(a^2 + b^2) - p^2} \]

where \( p = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \).