The value of the expression `(cos^(6)theta+sin^(6)theta-1)(tan^(2)theta+cot^(2)theta+2)` is :

Options

A. 1
B. -1
C. -3
D. 0

Answer

To solve the expression \((\cos^6 \theta + \sin^6 \theta - 1)(\tan^2 \theta + \cot^2 \theta + 2)\), we will break it down step by step.

### Step 1: Simplify \(\cos^6 \theta + \sin^6 \theta - 1\)

We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \(a = \cos^2 \theta\) and \(b = \sin^2 \theta\). Then:
\[ \cos^6 \theta + \sin^6 \theta = (\cos^2 \theta)^3 + (\sin^2 \theta)^3 = (\cos^2 \theta + \sin^2 \theta)((\cos^2 \theta)^2 - \cos^2 \theta \sin^2 \theta + (\sin^2 \theta)^2) \]
Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ \cos^6 \theta + \sin^6 \theta = 1 \cdot \left(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta\right) \]
Now, we can simplify \(\cos^4 \theta + \sin^4 \theta\):
\[ \cos^4 \theta + \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)^2 - 2\cos^2 \theta \sin^2 \theta = 1 - 2\cos^2 \theta \sin^2 \theta \]
Thus:
\[ \cos^6 \theta + \sin^6 \theta = (1 - 2\cos^2 \theta \sin^2 \theta) - \cos^2 \theta \sin^2 \theta = 1 - 3\cos^2 \theta \sin^2 \theta \]
So:
\[ \cos^6 \theta + \sin^6 \theta - 1 = -3\cos^2 \theta \sin^2 \theta \]

### Step 2: Simplify \(\tan^2 \theta + \cot^2 \theta + 2\)

Recall that:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}, \quad \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]
Thus:
\[ \tan^2 \theta + \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} \]
Finding a common denominator:
\[ \tan^2 \theta + \cot^2 \theta = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} \]
Using the earlier result:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Thus:
\[ \tan^2 \theta + \cot^2 \theta = \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]
Adding 2:
\[ \tan^2 \theta + \cot^2 \theta + 2 = \frac{1 - 2\sin^2 \theta \cos^2 \theta + 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1 + \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]

### Step 3: Combine the results

Now we can combine the two parts:
\[ (\cos^6 \theta + \sin^6 \theta - 1)(\tan^2 \theta + \cot^2 \theta + 2) = (-3\cos^2 \theta \sin^2 \theta) \left(\frac{1 + \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}\right) \]
This simplifies to:
\[ -3(1 + \sin^2 \theta \cos^2 \theta) = -3 - 3\sin^2 \theta \cos^2 \theta \]
Since \(\sin^2 \theta \cos^2 \theta\) is a positive term, the expression evaluates to:
\[ -3 \]

### Final Answer:
The value of the expression is \(-3\).