Question

Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

Hint:

When dealing with mixed \(\tan\) and \(\cot\) identities, converting everything to \(\tan \theta\) or \(\sin \theta, \cos \theta\) is usually the most effective strategy.

Answer 1

We need to prove:
\[ \frac{\tan\theta}{1 - \cot\theta} + \frac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta \]

Step 1: Rewrite cotθ as 1/tanθ
\[ \cot\theta = \frac{1}{\tan\theta} \]
Substitute this into the expression:
\[ \frac{\tan\theta}{1 - \frac{1}{\tan\theta}} + \frac{\frac{1}{\tan\theta}}{1 - \tan\theta} \]

Step 2: Simplify the first term
Denominator:
\[ 1 - \frac{1}{\tan\theta} = \frac{\tan\theta - 1}{\tan\theta} \] So the first term becomes:
\[ \frac{\tan\theta}{\frac{\tan\theta - 1}{\tan\theta}} = \frac{\tan^2\theta}{\tan\theta - 1} \]

Step 3: Simplify the second term
\[ \frac{\frac{1}{\tan\theta}}{1 - \tan\theta} = \frac{1}{\tan\theta(1 - \tan\theta)} \] Factor out -1:
\[ = -\frac{1}{\tan\theta(\tan\theta - 1)} \]

Step 4: Add the two terms
\[ \frac{\tan^2\theta}{\tan\theta - 1} - \frac{1}{\tan\theta(\tan\theta - 1)} \] Take common denominator \( \tan\theta(\tan\theta - 1) \):
\[ = \frac{\tan^3\theta - 1}{\tan\theta(\tan\theta - 1)} \]

Step 5: Use identity
\[ \tan^3\theta - 1 = (\tan\theta - 1)(\tan^2\theta + \tan\theta + 1) \]
Cancel \( \tan\theta - 1 \):
\[ = \frac{\tan^2\theta + \tan\theta + 1}{\tan\theta} \]
Separate the terms:
\[ = \frac{\tan^2\theta}{\tan\theta} + \frac{\tan\theta}{\tan\theta} + \frac{1}{\tan\theta} \] \[ = \tan\theta + 1 + \cot\theta \]

Final Answer:
\[ \boxed{ \frac{\tan\theta}{1 - \cot\theta} + \frac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta} \] Hence proved.