Question

Prove that: \[ \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \]

Answer 1

Step 1: Simplify $\frac{\tan \theta}{1 - \cot \theta}$
\[\frac{\tan \theta}{1 - \cot \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}}= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}.\]
Step 2: Simplify $\frac{\cot \theta}{1 - \tan \theta}$
\[\frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}= \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}= \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}.\]
Step 3: Add the two expressions
\[\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2\theta}{\sin \theta (\cos \theta - \sin \theta)}.\]
Simplify:
\[= \frac{\sin^3 \theta - \cos^3 \theta + \cos^3 \theta - \sin^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}.\]
Combine terms:
\[= 1 + \tan \theta + \cot \theta.\]