Question

Prove that: \[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \]

Answer 1

Step 1: Initial Expression and Goal:
Given expression:
\[\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}\]Target expression to prove:
\[1 + \tan \theta + \cot \theta\]

Step 2: Rewrite in terms of $\tan \theta$:
Substitute $\cot \theta = \frac{1}{\tan \theta}$ into the given expression:
\[\frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}\]Simplify denominators:
\[\frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}\]Simplify fractions:
\[\frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{\tan \theta (1 - \tan \theta)}\]

Step 3: Combine Fractions:
Rewrite the second term to match the denominator of the first:
\[\frac{1}{\tan \theta (1 - \tan \theta)} = \frac{-1}{\tan \theta (\tan \theta - 1)}\]Combine the terms with a common denominator:
\[\frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{\tan \theta (\tan \theta - 1)} = \frac{\tan^3 \theta - 1}{\tan \theta (\tan \theta - 1)}\]

Step 4: Factor and Simplify:
Factor the numerator (difference of cubes):
\[\tan^3 \theta - 1 = (\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)\]Substitute the factored numerator back into the expression:
\[\frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{\tan \theta (\tan \theta - 1)}\]Cancel $(\tan \theta - 1)$:
\[\frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta}\]Divide each term in the numerator by $\tan \theta$:
\[\frac{\tan^2 \theta}{\tan \theta} + \frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} = \tan \theta + 1 + \cot \theta\]

Step 5: Conclusion:
It has been demonstrated that:
\[\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\]The given identity is proven.