Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined: \(\frac{\text{tan θ}}{(\text{1 - cot θ})} + \frac{\text{cot θ}}{(\text{1 - tan θ})} = 1 + \text{secθ cosecθ}\) [Hint: Write the expression in terms of \(\text{sin θ}\) and \(\text{cos θ}\)]

Answer 1

\(\frac{\text{tan θ}}{(\text{1 - cot θ})} + \frac{\text{cot θ}}{(\text{1 - tan θ})} = 1 + \text{secθ cosecθ}\)

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).