Let `P(x)=((1-cos2x+sin2x)/(1+cos2x+s in2x))^2+((1+cotx+cot^2x)/(1+tanx+tan^2x)),` then the minimum value of `P(x)` equal 1 (b) 2 (c) 4 (d) 16

Options

A. 1
B. 2
C. 4
D. 16

Explanation

To find the minimum value of the function \( P(x) = \left( \frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} \right)^2 + \left( \frac{1 + \cot x + \cot^2 x}{1 + \tan x + \tan^2 x} \right) \), we will break it down into manageable steps.

### Step 1: Simplify the first term
We start with the first term:
\[ \frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} \]
Using the identities for \( \cos 2x \) and \( \sin 2x \):
\[ \cos 2x = 1 - 2\sin^2 x \quad \text{and} \quad \sin 2x = 2\sin x \cos x \]
Substituting these into the expression gives:
\[ 1 - (1 - 2\sin^2 x) + 2\sin x \cos x = 2\sin^2 x + 2\sin x \cos x \]
And for the denominator:
\[ 1 + (1 - 2\sin^2 x) + 2\sin x \cos x = 2 - 2\sin^2 x + 2\sin x \cos x \]
Thus, we have:
\[ \frac{2(\sin^2 x + \sin x \cos x)}{2(1 - \sin^2 x + \sin x \cos x)} = \frac{\sin^2 x + \sin x \cos x}{1 - \sin^2 x + \sin x \cos x} \]

### Step 2: Square the first term
Now we square this term:
\[ \left( \frac{\sin^2 x + \sin x \cos x}{1 - \sin^2 x + \sin x \cos x} \right)^2 \]

### Step 3: Simplify the second term
Next, we simplify the second term:
\[ \frac{1 + \cot x + \cot^2 x}{1 + \tan x + \tan^2 x} \]
Using the identities \( \cot x = \frac{\cos x}{\sin x} \) and \( \tan x = \frac{\sin x}{\cos x} \):
\[ 1 + \frac{\cos x}{\sin x} + \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^2 x + \cos x \sin x + \cos^2 x}{\sin^2 x} = \frac{1 + \sin x \cos x}{\sin^2 x} \]
And for the denominator:
\[ 1 + \frac{\sin x}{\cos x} + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x + \sin x \cos x + \sin^2 x}{\cos^2 x} = \frac{1 + \sin x \cos x}{\cos^2 x} \]
Thus, we have:
\[ \frac{\frac{1 + \sin x \cos x}{\sin^2 x}}{\frac{1 + \sin x \cos x}{\cos^2 x}} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \]

### Step 4: Combine the terms
Now we combine both terms:
\[ P(x) = \left( \frac{\sin^2 x + \sin x \cos x}{1 - \sin^2 x + \sin x \cos x} \right)^2 + \cot^2 x \]

### Step 5: Analyze the minimum value
From the AM-GM inequality, we know that:
\[ \tan^2 x + \cot^2 x \geq 2 \]
Thus, the minimum value of \( \tan^2 x + \cot^2 x \) is 2.

### Conclusion
Therefore, the minimum value of \( P(x) \) is:
\[ \text{Minimum value of } P(x) = 2 \]
Thus, the answer is option (b) 2.