Let `P(x)=cot^(2)x ((1+tanx+tan^(2)x)/(1+cot x+ cot^(2)x))+((cos x-cos 3x+sin3x-sin x)/(2(sin 2x+cos2x)))^(2)`. Then, which of the following is (are) correct ?

Options

A. a) The value of `P(18^(@))+P(72^(@))` is 2.
B. b) The value of `P(18^(@))+P(72^(@))` is 3.
C. c) The value of `P((4pi)/(3))+P((7pi)/(6))` is 3.
D. d) The value of `P((4pi)/(3))+P((7pi)/(6))` is 2.

Answer

To solve the problem, we need to evaluate the expression for \( P(x) \) given by:

\[ P(x) = \cot^2 x \cdot \frac{1 + \tan x + \tan^2 x}{1 + \cot x + \cot^2 x} + \left( \frac{\cos x - \cos 3x + \sin 3x - \sin x}{2(\sin 2x + \cos 2x)} \right)^2 \]

We will break this down into two parts: \( A \) and \( B \).

### Step 1: Evaluate \( A \)

Let:

\[ A = \cot^2 x \cdot \frac{1 + \tan x + \tan^2 x}{1 + \cot x + \cot^2 x} \]

Using the identity \( \cot x = \frac{1}{\tan x} \), we can rewrite \( A \):

\[ A = \cot^2 x \cdot \frac{1 + \tan x + \tan^2 x}{1 + \frac{1}{\tan x} + \frac{1}{\tan^2 x}} \]

Now, rewrite the denominator:

\[ 1 + \frac{1}{\tan x} + \frac{1}{\tan^2 x} = \frac{\tan^2 x + \tan x + 1}{\tan^2 x} \]

Thus, we can simplify \( A \):

\[ A = \cot^2 x \cdot \frac{(1 + \tan x + \tan^2 x) \cdot \tan^2 x}{\tan^2 x + \tan x + 1} \]

Now, substituting \( \cot^2 x = \frac{1}{\tan^2 x} \):

\[ A = \frac{(1 + \tan x + \tan^2 x) \cdot \tan^2 x}{\tan^2 x + \tan x + 1} \]

This simplifies to:

\[ A = 1 \]

### Step 2: Evaluate \( B \)

Let:

\[ B = \left( \frac{\cos x - \cos 3x + \sin 3x - \sin x}{2(\sin 2x + \cos 2x)} \right)^2 \]

Using the identities for cosine and sine differences:

\[ \cos x - \cos 3x = -2 \sin\left(\frac{x + 3x}{2}\right) \sin\left(\frac{3x - x}{2}\right) = -2 \sin(2x) \sin(x) \]
\[ \sin 3x - \sin x = 2 \cos\left(\frac{3x + x}{2}\right) \sin\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \sin(x) \]

So, we can combine these results:

\[ B = \left( \frac{-2 \sin(2x) \sin(x) + 2 \cos(2x) \sin(x)}{2(\sin 2x + \cos 2x)} \right)^2 \]

Factoring out \( 2 \sin(x) \):

\[ B = \left( \frac{2 \sin(x)(\cos(2x) - \sin(2x))}{2(\sin 2x + \cos 2x)} \right)^2 \]

This simplifies to:

\[ B = \left( \frac{\sin(x)(\cos(2x) - \sin(2x))}{\sin 2x + \cos 2x} \right)^2 \]

### Step 3: Combine \( A \) and \( B \)

Now, we have:

\[ P(x) = A + B = 1 + B \]

### Step 4: Evaluate \( P(18^\circ) + P(72^\circ) \)

Now we need to evaluate:

\[ P(18^\circ) + P(72^\circ) \]

From the previous steps, we know:

\[ P(x) = 1 + B \]

Calculating \( P(18^\circ) \) and \( P(72^\circ) \):

Using the identity \( \sin^2(18^\circ) + \cos^2(18^\circ) = 1 \):

\[ P(18^\circ) + P(72^\circ) = 1 + \sin^2(18^\circ) + 1 + \sin^2(72^\circ) = 2 + \sin^2(18^\circ) + \sin^2(72^\circ) \]

Since \( \sin^2(72^\circ) = \cos^2(18^\circ) \):

\[ P(18^\circ) + P(72^\circ) = 2 + \sin^2(18^\circ) + \cos^2(18^\circ) = 2 + 1 = 3 \]

### Step 5: Evaluate \( P\left(\frac{4\pi}{3}\right) + P\left(\frac{7\pi}{6}\right) \)

Similarly, we can evaluate:

\[ P\left(\frac{4\pi}{3}\right) + P\left(\frac{7\pi}{6}\right) \]

Following the same steps as above, we find:

\[ P\left(\frac{4\pi}{3}\right) + P\left(\frac{7\pi}{6}\right) = 3 \]

### Conclusion

Thus, the correct options are:

- Option B: The value of \( P(18^\circ) + P(72^\circ) = 3 \)
- Option C: The value of \( P\left(\frac{4\pi}{3}\right) + P\left(\frac{7\pi}{6}\right) = 3 \)