Question

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

Hint:

Expressing the entire expression in terms of $\tan\theta$ simplifies the algebra considerably compared to working with sines and cosines, which often leads to more complicated fraction handling.

Answer 1

Step 1: Understanding the Question:
We are given a trigonometric identity involving $\tan \theta$ and $\cot \theta$.
We need to prove that the Left-Hand Side (LHS) is equal to the Right-Hand Side (RHS).

Step 2: Key Formula or Approach:
To prove this identity, we can express all terms in the LHS in terms of $\tan \theta$ by using the reciprocal identity:
\[ \cot \theta = \frac{1}{\tan \theta} \]
We will also use the algebraic difference of cubes factorization identity:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

Step 3: Detailed Explanation:

• Write down the Left-Hand Side (LHS) of the expression:
\[ \text{LHS} = \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} \]

• Express $\cot \theta$ as $\frac{1}{\tan \theta}$:
\[ \text{LHS} = \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \]

• Simplify both terms:
- First term:
\[ \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} = \frac{\tan^2 \theta}{\tan \theta - 1} \]
- Second term:
\[ \frac{1}{\tan \theta (1 - \tan \theta)} = -\frac{1}{\tan \theta (\tan \theta - 1)} \]

• Combine both simplified terms over a common denominator:
\[ \text{LHS} = \frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{\tan \theta (\tan \theta - 1)} \]
\[ \text{LHS} = \frac{\tan^3 \theta - 1}{\tan \theta (\tan \theta - 1)} \]

• Factorize the numerator using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ where $a = \tan \theta$ and $b = 1$:
\[ \tan^3 \theta - 1 = (\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1) \]

• Substitute this factorization back into the expression:
\[ \text{LHS} = \frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{\tan \theta (\tan \theta - 1)} \]
Cancel the common factor $(\tan \theta - 1)$ from the numerator and denominator:
\[ \text{LHS} = \frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta} \]

• Split the fraction to match the RHS format:
\[ \text{LHS} = \frac{\tan^2 \theta}{\tan \theta} + \frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} \]
\[ \text{LHS} = \tan \theta + 1 + \cot \theta \]
Rearrange the terms:
\[ \text{LHS} = 1 + \tan \theta + \cot \theta = \text{RHS} \]


Step 4: Final Answer:
Since the Left-Hand Side simplifies exactly to the Right-Hand Side, the identity is proven.