\(\frac{\text{tan θ}}{(\text{1 - cot θ})} + \frac{\text{cot θ}}{(\text{1 - tan θ})} = 1 + \text{secθ cosecθ}\)
Left Hand Side (LHS): \(=\frac{\text{tan θ}}{(\text{1 - cot θ})} + \frac{\text{cot θ}}{(\text{1 - tan θ})}\)
Expressing in terms of sine and cosine:
\(= \frac{(\frac{\text{sin θ}}{\text{cos θ}})}{\left(1 - \frac{\text{cos θ}}{\text{sin θ}}\right)} + \frac{(\frac{\text{cos θ}}{\text{sin θ}})}{1 - \frac{\text{cos θ}}{\text{sin θ}}}\)
\(= \frac{(\frac{\text{sin θ}}{\text{cos θ}})}{\frac{(\text{sin θ - cos θ})}{\text{sin θ}}} + \frac{(\frac{\text{cos θ}}{\text{sin θ}})}{\frac{(\text{cos θ - sin θ})}{\text{cos θ}}}\)
\(= \frac{\text{sin² θ}}{\text{cos θ(sin θ - cos θ})} + \frac{\text{cos² θ}}{\text{sin θ(sin θ - cos θ})}\)
Common factor \(\frac{1}{(\text{sin θ - cos θ})}\) taken out:
\(=\frac{1}{(\text{sin θ - cos θ})}\) \(\left[\frac{\text{sin²θ}}{\text{cos θ}} - \frac{\text{cos²θ}}{\text{sin θ}}\right]\)
Combining terms within the bracket:
\(=\frac{1}{(\text{sin θ - cos θ})}\)\([\frac{(\text{sin³θ - cos³θ})}{\text{sin θ cos θ}}]\)
Applying the difference of cubes identity (a³ - b³ = (a - b)(a² + ab + b²)):
\(=\frac{1}{(\text{sin θ - cos θ})}\) \([\frac{((\text{sin θ - cos θ) sin²θ+ cos²θ + sin θcos θ})}{\text{sin θ cos θ}}]\)
Simplifying using the identity sin² A + cos² A = 1:
\(= \frac{(1 + \text{sin θ cos θ})}{(\text{sin θ cos θ})} \)
Separating terms:
\(= 1 + \text{sec θ cosec θ}\)
= Right Hand Side (RHS).