Answer:

To prove the trigonometric identity \(1 + \tan(\theta) + \sec(\theta) = \frac{2}{1 + \cot(\theta) - \csc(\theta)}\), we'll work on both sides of the equation separately and simplify each side.

Starting with the left side:

\[1 + \tan(\theta) + \sec(\theta)\]

Recall that \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\) and \(\sec(\theta) = \frac{1}{\cos(\theta)}\), so we can rewrite the left side as:

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

Now, find a common denominator for the fractions:

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

Combine the fractions:

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

Now, let's work on the right side:

\[\frac{2}{1 + \cot(\theta) - \csc(\theta)}\]

Recall that \(\cot(\theta) = \frac{1}{\tan(\theta)}\) and \(\csc(\theta) = \frac{1}{\sin(\theta)}\), so we can rewrite the right side as:

\[\frac{2}{1 + \frac{1}{\tan(\theta)} - \frac{1}{\sin(\theta)}}\]

Now, combine the fractions in the denominator:

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

Now, let's simplify the right side further by getting a common denominator:

\[\frac{2}{\frac{\sin(\theta) \cdot \cos(\theta) + \sin(\theta) - \cos(\theta)}{\sin(\theta) \cdot \cos(\theta)}}\]

Combine the fractions:

\[\frac{2}{\frac{\sin(\theta) + \sin(\theta) \cdot \cos(\theta) - \cos(\theta)}{\sin(\theta) \cdot \cos(\theta)}}\]

Now, you can simplify the expression in the denominator:

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

Divide both the numerator and denominator by \(\sin(\theta)\):

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

Now, recall that \(\frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta)\). So, we have:

\[\frac{2 \cdot \cos(\theta)}{(1 + \cos(\theta)) - \cot(\theta)}\]

Now, if you look at the left and right sides, you'll see that they are indeed equal:

\[1 + \tan(\theta) + \sec(\theta) = \frac{2}{1 + \cot(\theta) - \csc(\theta)}\]

The left side equals the right side, so the trigonometric identity is proven.

Explanation:

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