Question

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

Answer 1

Step 1: Start with the given expression:
We are given the expression:
\[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} \] We need to prove that this expression equals:
\[ 1 + \tan \theta + \cot \theta \]

Step 2: Express $\cot \theta$ in terms of $\tan \theta$:
Recall that $\cot \theta = \frac{1}{\tan \theta}$. Using this, we can rewrite the given expression:
\[ \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \] Simplify the denominators:
\[ \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \] Now simplify each fraction:
\[ \frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{\tan \theta (1 - \tan \theta)} \]

Step 3: Combine the terms:
Now, observe that the two terms have a similar denominator. We can rewrite the second term as:
\[ \frac{1}{\tan \theta (1 - \tan \theta)} = \frac{-1}{\tan \theta (\tan \theta - 1)} \] Thus, the expression becomes:
\[ \frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{\tan \theta (\tan \theta - 1)} \] Now, we combine the terms by writing them over a common denominator:
\[ \frac{\tan^3 \theta - 1}{\tan \theta (\tan \theta - 1)} \]

Step 4: Simplify the numerator:
Notice that the numerator $\tan^3 \theta - 1$ is a difference of cubes, which can be factored as:
\[ \tan^3 \theta - 1 = (\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1) \] Thus, the expression becomes:
\[ \frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{\tan \theta (\tan \theta - 1)} \] Cancel the common factor of $(\tan \theta - 1)$ in the numerator and denominator:
\[ \frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta} \] Now, divide each term in the numerator by $\tan \theta$:
\[ \frac{\tan^2 \theta}{\tan \theta} + \frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} \] This simplifies to:
\[ \tan \theta + 1 + \cot \theta \]

Step 5: Final conclusion:
We have shown that:
\[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \] Thus, the given expression is true.