Find the value of \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}\).
tan θ + sec θ
Divide the numerator and the denominator by \(\cos \theta\): the expression becomes \(\frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}\).
Use the identity \(\sec^2 \theta - \tan^2 \theta = 1\), i.e. \(1 = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)\), to replace the 1 in the numerator.
Numerator = \((\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = (\sec \theta + \tan \theta)(1 - \sec \theta + \tan \theta)\).
The denominator is \(\tan \theta + 1 - \sec \theta = 1 - \sec \theta + \tan \theta\), which cancels, so the expression equals \(\tan \theta + \sec \theta\).
Check: at \(\theta = 60^{\circ}\), the expression is \(\frac{\frac{\sqrt{3}}{2} - \frac{1}{2} + 1}{\frac{\sqrt{3}}{2} + \frac{1}{2} - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}\), and \(\tan 60^{\circ} + \sec 60^{\circ} = \sqrt{3} + 2\), the same value.
Hence, tan θ + sec θ.
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