Prove that tan θ /(1 - cot θ) + cot θ / (1 - tan θ) = 1 + sec θ cosecθ

Q) Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

Ans: Let’s start from LHS

$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

= $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$

= $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$

= $(\frac{1}{\sin \theta - \cos \theta}) (\frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta})$

= $(\frac{1}{\sin \theta - \cos \theta}) (\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta})$

We know that, a³−b³ formula is = (a−b)(a² + b² + ab)

$\therefore (\frac{1}{\sin \theta - \cos \theta}) (\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos ^2 \theta + \sin \theta cos\theta)}{\sin \theta \cos \theta})$