Prove that `cot^4 theta+ cot^2 theta=` `cosec^4 theta-cosec^2 theta`

Answer

To prove that \( \cot^4 \theta + \cot^2 \theta = \csc^4 \theta - \csc^2 \theta \), we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Start with the LHS
We begin with the left-hand side:
\[ \text{LHS} = \cot^4 \theta + \cot^2 \theta \]

### Step 2: Factor out \( \cot^2 \theta \)
We can factor out \( \cot^2 \theta \) from the expression:
\[ \text{LHS} = \cot^2 \theta (\cot^2 \theta + 1) \]

### Step 3: Use the identity for \( \cot^2 \theta + 1 \)
Using the trigonometric identity \( \cot^2 \theta + 1 = \csc^2 \theta \), we can substitute:
\[ \text{LHS} = \cot^2 \theta \cdot \csc^2 \theta \]

### Step 4: Rewrite \( \cot^2 \theta \) in terms of \( \csc^2 \theta \)
We know that \( \cot^2 \theta = \csc^2 \theta - 1 \). Therefore, we can substitute \( \cot^2 \theta \):
\[ \text{LHS} = (\csc^2 \theta - 1) \cdot \csc^2 \theta \]

### Step 5: Expand the expression
Now, we expand the expression:
\[ \text{LHS} = \csc^4 \theta - \csc^2 \theta \]

### Step 6: Conclude that LHS equals RHS
Thus, we have shown that:
\[ \text{LHS} = \csc^4 \theta - \csc^2 \theta = \text{RHS} \]

### Final Result
We have proved that:
\[ \cot^4 \theta + \cot^2 \theta = \csc^4 \theta - \csc^2 \theta \]