Show that `((1 + cos theta - sin ^(2) theta)/( sin theta (1 + cos theta ))) = cot theta `

Answer

To show that

\[ \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta, \]

we will simplify the left-hand side (LHS) step by step.

### Step 1: Write down the LHS
The left-hand side of the equation is:

\[ \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)}. \]

### Step 2: Use the Pythagorean Identity
Recall the Pythagorean identity:

\[ \sin^2 \theta + \cos^2 \theta = 1. \]

From this, we can express \(\sin^2 \theta\) as:

\[ \sin^2 \theta = 1 - \cos^2 \theta. \]

Substituting this into the LHS gives:

\[ 1 + \cos \theta - (1 - \cos^2 \theta) = 1 + \cos \theta - 1 + \cos^2 \theta = \cos \theta + \cos^2 \theta. \]

### Step 3: Rewrite the LHS
Now, we can rewrite the LHS as:

\[ \frac{\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}. \]

### Step 4: Factor out \(\cos \theta\)
Notice that we can factor \(\cos \theta\) from the numerator:

\[ \frac{\cos \theta (1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}. \]

### Step 5: Cancel out \(1 + \cos \theta\)
Assuming \(1 + \cos \theta \neq 0\), we can cancel \(1 + \cos \theta\) from the numerator and the denominator:

\[ \frac{\cos \theta}{\sin \theta}. \]

### Step 6: Write in terms of cotangent
The expression \(\frac{\cos \theta}{\sin \theta}\) is equal to \(\cot \theta\):

\[ \cot \theta. \]

### Conclusion
Thus, we have shown that:

\[ \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta, \]

which confirms that the left-hand side is equal to the right-hand side (RHS).

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