Simplifying the Trigonometric Expression

We need to find the value of the given trigonometric expression:

\(\frac{\sin \theta + \cos \theta - 1}{\sin \theta - \cos \theta + 1} \times \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}\)

Let's simplify the expression by tackling each part separately.

Step 1: Simplify the Square Root Term

Consider the term under the square root:

\(\sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}\)

To rationalize the denominator inside the square root, we multiply the numerator and denominator by \((1 + \sin \theta)\):

\(\sqrt{\frac{(1 + \sin \theta)(1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}} = \sqrt{\frac{(1 + \sin \theta)^2}{1^2 - \sin^2 \theta}}\)

Using the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), we know that \(1 - \sin^2 \theta = \cos^2 \theta\). Substitute this into the expression:

\(\sqrt{\frac{(1 + \sin \theta)^2}{\cos^2 \theta}}\)

Assuming \(\cos \theta > 0\) and \(1 + \sin \theta \ge 0\) (which is generally true for the domain where the original expression is defined), the square root simplifies to:

\(\frac{1 + \sin \theta}{\cos \theta}\)

Step 2: Simplify the Fractional Term

Now let's simplify the first fraction:

\(\frac{\sin \theta + \cos \theta - 1}{\sin \theta - \cos \theta + 1}\)

We can group terms and use the difference of squares identity. Rewrite the numerator as \((\sin \theta + \cos \theta) - 1\) and the denominator as \((\sin \theta + 1) - \cos \theta\). A useful technique is to multiply the numerator and denominator by \((\sin \theta + \cos \theta + 1)\) (which is similar to rationalizing but for trigonometric terms):

\(\frac{(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)}{(\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)}\)

In the numerator, use the identity \((a-b)(a+b) = a^2 - b^2\) with \(a = \sin \theta + \cos \theta\) and \(b = 1\):

Numerator: \((\sin \theta + \cos \theta)^2 - 1^2 = (\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta) - 1\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

Numerator: \((1 + 2\sin \theta \cos \theta) - 1 = 2\sin \theta \cos \theta\)

In the denominator, group terms as \((\sin \theta + 1 - \cos \theta)(\sin \theta + 1 + \cos \theta)\). Use the identity \((a-b)(a+b) = a^2 - b^2\) with \(a = \sin \theta + 1\) and \(b = \cos \theta\):

Denominator: \((\sin \theta + 1)^2 - \cos^2 \theta = (\sin^2 \theta + 2\sin \theta + 1) - \cos^2 \theta\)

Using \(1 - \cos^2 \theta = \sin^2 \theta\):

Denominator: \(\sin^2 \theta + 2\sin \theta + \sin^2 \theta = 2\sin^2 \theta + 2\sin \theta = 2\sin \theta (\sin \theta + 1)\)

So, the first fraction simplifies to:

\(\frac{2\sin \theta \cos \theta}{2\sin \theta (\sin \theta + 1)}\)

Assuming \(\sin \theta \ne 0\) and \(\sin \theta + 1 \ne 0\), we can cancel the \(2\sin \theta\) term:

\(\frac{\cos \theta}{\sin \theta + 1}\)

Step 3: Multiply the Simplified Terms

Now we multiply the simplified square root term from Step 1 and the simplified fractional term from Step 2:

\(\left(\frac{\cos \theta}{\sin \theta + 1}\right) \times \left(\frac{1 + \sin \theta}{\cos \theta}\right)\)

Assuming \(\cos \theta \ne 0\) and \(1 + \sin \theta \ne 0\), we can cancel the common terms \(\cos \theta\) and \((1 + \sin \theta)\):

\(\frac{\cos \theta}{1 + \sin \theta} \times \frac{1 + \sin \theta}{\cos \theta} = 1\)

Thus, the value of the given trigonometric expression is 1.

Revision Table: Important Trigonometric Identities

This solution heavily relied on fundamental trigonometric identities and algebraic manipulations. Key identities used include:

• Pythagorean Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Rearrangements of Pythagorean Identity: \(1 - \sin^2 \theta = \cos^2 \theta\) and \(1 - \cos^2 \theta = \sin^2 \theta\)
• Algebraic Identity: \(a^2 - b^2 = (a-b)(a+b)\)
• Algebraic Identity: \((a+b)^2 = a^2 + 2ab + b^2\)

Additional Information: Conditions for the Expression to be Defined

For the original trigonometric expression to be valid, several conditions must be met regarding the value of \(\theta\):

• The denominator of the first fraction cannot be zero: \(\sin \theta - \cos \theta + 1 \ne 0\).
• The term under the square root must be non-negative: \(\frac{1 + \sin \theta}{1 - \sin \theta} \ge 0\). Since \(1 + \sin \theta \ge 0\) for all real \(\theta\), this implies \(1 - \sin \theta > 0\) (the denominator cannot be zero), which means \(\sin \theta < 1\).
• The denominator in the simplified square root term, \(\cos \theta\), cannot be zero. Thus, \(\cos \theta \ne 0\).
• The denominator in the simplified first fraction, \(\sin \theta + 1\), cannot be zero. Thus, \(\sin \theta \ne -1\). This condition \(\sin \theta \ne -1\) is already covered by the condition \(\sin \theta < 1\).

So, for the expression to be defined and for our simplifications involving division and cancellation to be valid, we generally need \(\sin \theta < 1\) and \(\cos \theta \ne 0\). Also, the simplification of the square root \(\sqrt{\cos^2 \theta}\) to \(\cos \theta\) assumed \(\cos \theta > 0\). If \(\cos \theta < 0\), the simplified square root is \(\frac{1 + \sin \theta}{-\cos \theta}\), and the final product would be \(\frac{\cos \theta}{\sin \theta + 1} \times \frac{1 + \sin \theta}{-\cos \theta} = -1\). However, given the options, the problem likely assumes conditions where the value is 1, such as \(\theta\) being in the first quadrant.