To prove the identity [tex]\(\frac{1 - \cos^4 \theta}{\sin^4 \theta} = 1 + 2 \cot^2 \theta\)[/tex]:
1. Simplify the Left-Hand Side (LHS):
[tex]\[\frac{1 - \cos^4 \theta}{\sin^4 \theta}\][/tex]
We start by expressing [tex]\(1 - \cos^4 \theta\)[/tex] in a factorized form.
Notice that:
[tex]\[1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta)\][/tex]
Therefore:
[tex]\[\frac{1 - \cos^4 \theta}{\sin^4 \theta} = \frac{(1 - \cos^2 \theta)(1 + \cos^2 \theta)}{\sin^4 \theta}\][/tex]
2. Use the Pythagorean identity:
[tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
Therefore, [tex]\(1 - \cos^2 \theta = \sin^2 \theta\)[/tex].
Substituting this in:
[tex]\[\frac{(\sin^2 \theta)(1 + \cos^2 \theta)}{\sin^4 \theta}\][/tex]
Simplify by cancelling [tex]\(\sin^2 \theta\)[/tex] from the numerator and the denominator:
[tex]\[\frac{1 + \cos^2 \theta}{\sin^2 \theta}\][/tex]
3. Rewrite the expression:
[tex]\(\frac{1 + \cos^2 \theta}{\sin^2 \theta}\)[/tex] can be broken down into:
[tex]\[\frac{1}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta}\][/tex]
Which simplifies to:
[tex]\(\csc^2 \theta + \cot^2 \theta\)[/tex].
4. Use another Pythagorean identity:
[tex]\(\csc^2 \theta = 1 + \cot^2 \theta\)[/tex].
So:
[tex]\[\csc^2 \theta + \cot^2 \theta = (1 + \cot^2 \theta) + \cot^2 \theta\][/tex]
[tex]\[= 1 + 2 \cot^2 \theta\][/tex]
5. Conclusion:
Therefore:
[tex]\[\frac{1 - \cos^4 \theta}{\sin^4 \theta} = 1 + 2 \cot^2 \theta\][/tex]
Thus, we have shown that the original identity holds true.
