Prove the identity \(\frac{1}{\cos \theta} - \frac{\cos \theta}{1 + \sin \theta} \equiv \tan \theta\).
Solution
Let \(s = \sin \theta\) and \(c = \cos \theta\).
The left-hand side (LHS) is:
\(\frac{1}{c} - \frac{c}{1+s}\)
Combine the fractions:
\(\frac{1 + s - c^2}{c(1+s)}\)
Using the identity \(s^2 + c^2 = 1\), we have \(c^2 = 1 - s^2\).
Substitute \(c^2\) in the expression:
\(\frac{1 + s - (1 - s^2)}{c(1+s)} = \frac{s^2 + s}{c(1+s)}\)
Factor out \(s\):
\(\frac{s(s + 1)}{c(1+s)}\)
Cancel \(1+s\) from numerator and denominator:
\(\frac{s}{c}\)
Thus, the LHS simplifies to \(\tan \theta\), proving the identity.
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