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

Step 1: Understanding the Concept:
We convert all terms into \(\tan \theta\) to simplify the expression on the L.H.S. and then use algebraic identities to reach the R.H.S.
Step 2: Detailed Explanation:
L.H.S. \(= \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}\)
Substitute \(\cot \theta = \frac{1}{\tan \theta}\):
\[ = \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \]
\[ = \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} + \frac{1}{\tan \theta (1 - \tan \theta)} \]
\[ = \frac{\tan^{2} \theta}{\tan \theta - 1} - \frac{1}{\tan \theta (\tan \theta - 1)} \]
Take L.C.M. \(\tan \theta (\tan \theta - 1)\):
\[ = \frac{\tan^{3} \theta - 1}{\tan \theta (\tan \theta - 1)} \]
Using identity \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\):
\[ = \frac{(\tan \theta - 1)(\tan^{2} \theta + \tan \theta + 1)}{\tan \theta (\tan \theta - 1)} \]
\[ = \frac{\tan^{2} \theta + \tan \theta + 1}{\tan \theta} \]
Divide each term in the numerator by \(\tan \theta\):
\[ = \frac{\tan^{2} \theta}{\tan \theta} + \frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} \]
\[ = \tan \theta + 1 + \cot \theta \]
L.H.S. \(=\) R.H.S. Hence Proved.
Step 3: Final Answer:
The trigonometric identity is proved.