Question

establish the identity
\\(\frac{1 - \sin\theta}{1 + \sin\theta}=(\tan\theta - \sec\theta)^2\\)

starting with the right side, which shows the key steps in establishing the identity?
\\(\circ a.\\)
\\((\tan\theta - \sec\theta)^2=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\frac{2\sin\theta}{\cos^{2}\theta}+\frac{1}{\cos^{2}\theta}=\frac{(\sin\theta - 1)^2}{1 - \cos^{2}\theta}=\frac{1 - \sin\theta}{1 + \sin\theta}\\)
\\(\circ b.\\)
\\((\tan\theta - \sec\theta)^2=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\frac{2\sin\theta}{\cos^{2}\theta}+\frac{1}{\cos^{2}\theta}=\frac{(\sin\theta - 1)^2}{1 - \sin^{2}\theta}=\frac{1 - \sin\theta}{1 + \sin\theta}\\)
\\(\circ c.\\)
\\((\tan\theta - \sec\theta)^2=\tan^{2}\theta - \sec^{2}\theta=\frac{(\sin\theta - 1)^2}{1 - \sin^{2}\theta}=\frac{1 - \sin\theta}{1 + \sin\theta}\\)
\\(\circ d.\\)
\\((\tan\theta - \sec\theta)^2=\tan^{2}\theta - \sec^{2}\theta=\frac{(\sin\theta - 1)^2}{1 - \cos^{2}\theta}=\frac{1 - \sin\theta}{1 + \sin\theta}\\)

Explanation:

Step1: Expand right - hand side

Recall that \((a - b)^2=a^{2}-2ab + b^{2}\). For \((\tan\theta-\sec\theta)^{2}\), since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\sec\theta=\frac{1}{\cos\theta}\), we have \((\tan\theta - \sec\theta)^{2}=\tan^{2}\theta-2\tan\theta\sec\theta+\sec^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\frac{2\sin\theta}{\cos^{2}\theta}+\frac{1}{\cos^{2}\theta}=\frac{\sin^{2}\theta - 2\sin\theta + 1}{\cos^{2}\theta}\).

Step2: Factor the numerator

Factor \(\sin^{2}\theta - 2\sin\theta + 1\) as \((\sin\theta - 1)^{2}\). So we get \(\frac{(\sin\theta - 1)^{2}}{\cos^{2}\theta}\).

Step3: Use the Pythagorean identity

Recall that \(\cos^{2}\theta=1 - \sin^{2}\theta=(1+\sin\theta)(1 - \sin\theta)\) by the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\). Then \(\frac{(\sin\theta - 1)^{2}}{\cos^{2}\theta}=\frac{(1-\sin\theta)^{2}}{(1 + \sin\theta)(1 - \sin\theta)}=\frac{1-\sin\theta}{1+\sin\theta}\).

Answer:

B. \((\tan\theta-\sec\theta)^{2}=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\frac{2\sin\theta}{\cos^{2}\theta}+\frac{1}{\cos^{2}\theta}=\frac{(\sin\theta - 1)^{2}}{1-\sin^{2}\theta}=\frac{1-\sin\theta}{1+\sin\theta}\)