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

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

LHS $=\frac{\text{tan θ}}{(\text{1 - cot θ})} + \frac{\text{cot θ}}{(\text{1 - tan θ})}$

$= \frac{(\frac{\text{sin θ}}{\text{cos θ}})}{\left(1 - \frac{\text{cos θ}}{\text{sin θ}}\right)} + \frac{(\frac{\text{cos θ}}{\text{sin θ}})}{1 - \frac{\text{cos θ}}{\text{sin θ}}}$

$= \frac{(\frac{\text{sin θ}}{\text{cos θ}})}{\frac{(\text{sin θ - cos θ})}{\text{sin θ}}} + \frac{(\frac{\text{cos θ}}{\text{sin θ}})}{\frac{(\text{cos θ - sin θ})}{\text{cos θ}}}$

$= \frac{\text{sin² θ}}{\text{cos θ(sin θ - cos θ})} + \frac{\text{cos² θ}}{\text{sin θ(sin θ - cos θ})}$

Taking $\frac{1}{(\text{sin θ - cos θ})}$ as common

$=\frac{1}{(\text{sin θ - cos θ})}$ $\left[\frac{\text{sin²θ}}{\text{cos θ}} - \frac{\text{cos²θ}}{\text{sin θ}}\right]$

=\frac{1}{(\text{sin θ - cos θ})}$$[\frac{(\text{sin³θ - cos³θ})}{\text{sin θ cos θ}}]

$=\frac{1}{(\text{sin θ - cos θ})}$ $[\frac{((\text{sin θ - cos θ) sin²θ+ cos²θ + sin θcos θ})}{\text{sin θ cos θ}}]$ [a³ - b³ = (a - b)(a² + ab + b²)]

$= \frac{(1 + \text{sin θ cos θ})}{(\text{sin θ cos θ})}$ (By Identity sin² A+ cos² A = 1)

$= 1 + \text{sec θ cosec θ}$
= R.H.S.