Problem Statement
Prove the following trigonometric identity:
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
Verified Solution & Marking Scheme
Convert All Ratios into sin θ and cos θ
$\text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}$
Factor out Common Term (sin θ - cos θ)
Since $(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta)$:
$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}$
$= \frac{1}{\sin \theta - \cos \theta} \left( \frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right)$
$= \frac{1}{\sin \theta - \cos \theta} \left( \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta} \right)$
Apply a³ - b³ Factoring Identity
Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:
$\sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$
$= (\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)$
Substituting this back:
$= \frac{(\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)}{(\sin \theta - \cos \theta)(\sin \theta \cos \theta)}$
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \sec \theta \csc \theta + 1$
$= 1 + \sec \theta \csc \theta = \text{RHS}$.