Question

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

Answer 1

Step 1: State the equation to be proven:
The identity to be proven is:
\[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \]

Step 2: Rewrite terms using the identity \( \cot \theta = \frac{1}{\tan \theta} \):
Begin by expressing the left-hand side (LHS) entirely in terms of \( \tan \theta \):
\[ \text{LHS} = \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \]
Simplify each term.

Step 3: Simplify the first term:
The first term is \( \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} \). Simplify the denominator:
\[ 1 - \frac{1}{\tan \theta} = \frac{\tan \theta - 1}{\tan \theta} \]
Substitute this back into the first term:
\[ \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} = \frac{\tan^2 \theta}{\tan \theta - 1} \]

Step 4: Simplify the second term:
The second term is \( \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \). Simplify as follows:
\[ \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} = \frac{1}{\tan \theta (1 - \tan \theta)} \]

Step 5: Combine the simplified terms:
Add the two simplified terms:
\[ \text{LHS} = \frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{\tan \theta (1 - \tan \theta)} \]
To combine them, rewrite the second term with a denominator of \( \tan \theta - 1 \):
\[ \frac{1}{\tan \theta (1 - \tan \theta)} = \frac{-1}{\tan \theta (\tan \theta - 1)} \]
The expression now is:
\[ \text{LHS} = \frac{\tan^2 \theta - 1}{\tan \theta - 1} \]
Factor the numerator using the difference of squares identity \( a^2 - b^2 = (a-b)(a+b) \), where \( a = \tan \theta \) and \( b = 1 \):
\[ \text{LHS} = \frac{(\tan \theta - 1)(\tan \theta + 1)}{\tan \theta - 1} \]
Cancel the common factor \( \tan \theta - 1 \) (provided \( \tan \theta eq 1 \), i.e., \( \theta eq 45^\circ \)):
\[ \text{LHS} = \tan \theta + 1 \]

Step 6: Finalize the proof:
The LHS simplifies to \( 1 + \tan \theta \). This is equivalent to the RHS, \( 1 + \tan \theta + \cot \theta \), only if \( \cot \theta = 0 \), which is not generally true. There appears to be an error in the provided steps or the original identity. However, following the steps as given, the LHS simplifies to \( 1 + \tan \theta \).
Assuming the provided steps are intended to reach the stated RHS, let's re-examine Step 5's combination: \[ \text{LHS} = \frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{\tan \theta (1 - \tan \theta)} = \frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{\tan \theta (\tan \theta - 1)} \] To combine these, the common denominator should be \( \tan \theta (\tan \theta - 1) \): \[ \text{LHS} = \frac{\tan^2 \theta \cdot \tan \theta}{(\tan \theta - 1) \cdot \tan \theta} - \frac{1}{\tan \theta (\tan \theta - 1)} = \frac{\tan^3 \theta - 1}{\tan \theta (\tan \theta - 1)} \] Using the difference of cubes identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \): \[ \text{LHS} = \frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{\tan \theta (\tan \theta - 1)} \] Cancel \( \tan \theta - 1 \): \[ \text{LHS} = \frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta} = \frac{\tan^2 \theta}{\tan \theta} + \frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} = \tan \theta + 1 + \cot \theta \]
This corrected combination shows that the LHS equals \( 1 + \tan \theta + \cot \theta \). Hence, the identity is proven: \[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \]