What is tan²α equal to?

Calculation: 

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We need to find \(\tan^2(\alpha) \) in terms of p  and q .

We start by simplifying the expression \(\frac{p}{p + 2q} \)

\( \frac{p}{p + 2q} = \frac{1}{1 + \frac{2q}{p}} \)

This simplifies further as:

\( \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} \)

Now, simplify the fraction inside the denominator:

\( = \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} = \frac{1}{1 + \frac{1}{2} \cdot \tan(\alpha) \cdot \tan(2\alpha)} \)

Now, expand both terms using trigonometric identities:

\( = \frac{\sin(\alpha) \cdot \sin(2\alpha)}{\sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \cos(2\alpha)} \)

Simplifying this expression gives:

\( = \tan^2(\alpha) \)

∴ The correct answer is Option (c):