`cosec^(6)theta-cot^(6)theta-3cot^(2)theta.cosec^(2)theta=?`

Options

A. 2
B. `-1`
C. 1
D. 4

Answer

To solve the expression \( \csc^6 \theta - \cot^6 \theta - 3 \cot^2 \theta \csc^2 \theta \), we can use the identity for the difference of cubes and some trigonometric identities. Let's break it down step by step:

### Step 1: Recognize the Difference of Cubes
We can rewrite \( \csc^6 \theta - \cot^6 \theta \) as a difference of cubes:
\[ \csc^6 \theta - \cot^6 \theta = (\csc^2 \theta)^3 - (\cot^2 \theta)^3 \]
Using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), we set \( a = \csc^2 \theta \) and \( b = \cot^2 \theta \).

### Step 2: Apply the Difference of Cubes Formula
Applying the formula:
\[ \csc^6 \theta - \cot^6 \theta = (\csc^2 \theta - \cot^2 \theta)(\csc^4 \theta + \csc^2 \theta \cot^2 \theta + \cot^4 \theta) \]

### Step 3: Simplify \( \csc^2 \theta - \cot^2 \theta \)
Using the identity \( \csc^2 \theta = 1 + \cot^2 \theta \):
\[ \csc^2 \theta - \cot^2 \theta = 1 \]

### Step 4: Substitute Back into the Expression
Now substituting back into our expression:
\[ \csc^6 \theta - \cot^6 \theta = 1 \cdot (\csc^4 \theta + \csc^2 \theta \cot^2 \theta + \cot^4 \theta) = \csc^4 \theta + \csc^2 \theta \cot^2 \theta + \cot^4 \theta \]

### Step 5: Combine with the Remaining Terms
Now we substitute this back into the original expression:
\[ \csc^4 \theta + \csc^2 \theta \cot^2 \theta + \cot^4 \theta - 3 \cot^2 \theta \csc^2 \theta \]
This simplifies to:
\[ \csc^4 \theta + (\csc^2 \theta \cot^2 \theta - 3 \cot^2 \theta \csc^2 \theta) + \cot^4 \theta \]
\[ = \csc^4 \theta - 2 \cot^2 \theta \csc^2 \theta + \cot^4 \theta \]

### Step 6: Factor the Expression
Notice that:
\[ \csc^4 \theta - 2 \cot^2 \theta \csc^2 \theta + \cot^4 \theta = (\csc^2 \theta - \cot^2 \theta)^2 \]
Since we already established that \( \csc^2 \theta - \cot^2 \theta = 1 \):
\[ (\csc^2 \theta - \cot^2 \theta)^2 = 1^2 = 1 \]

### Final Answer
Thus, the final answer is:
\[ \csc^6 \theta - \cot^6 \theta - 3 \cot^2 \theta \csc^2 \theta = 1 \]