Simplifying the Trigonometric Expression

We are asked to simplify the given trigonometric expression:

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

To simplify this expression, we can use algebraic manipulation and trigonometric identities. One common technique is to multiply the numerator and the denominator by a suitable expression that helps utilize identities like \(\sin^2 \theta + \cos^2 \theta = 1\) or \(\sec^2 \theta - \tan^2 \theta = 1\).

Let's multiply the numerator and the denominator by \((\sin \theta + \cos \theta + 1)\). This expression is chosen strategically because the denominator of the original expression is \((\sin \theta + \cos \theta) - 1\), and multiplying by \((\sin \theta + \cos \theta) + 1\) forms a difference of squares pattern.

The expression becomes:

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

Let's calculate the numerator and the denominator separately.

Numerator:

\((\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)\)

We can group terms as \((\sin \theta + 1 - \cos \theta)(\sin \theta + 1 + \cos \theta)\). This is in the form \((A-B)(A+B) = A^2 - B^2\), where \(A = \sin \theta + 1\) and \(B = \cos \theta\).

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

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

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(1 - \cos^2 \theta = \sin^2 \theta\). Also, \(1 = \sin^2 \theta + \cos^2 \theta\).

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

\( = 2\sin^2 \theta + 2\sin \theta\)

\( = 2\sin \theta (\sin \theta + 1)\)

Denominator:

\((\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)\)

This is in the form \((C-D)(C+D) = C^2 - D^2\), where \(C = \sin \theta + \cos \theta\) and \(D = 1\).

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

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

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

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

\( = 2\sin \theta \cos \theta\)

Now, substitute the simplified numerator and denominator back into the expression:

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

Assuming \(\sin \theta \neq 0\), we can cancel the \(2\sin \theta\) term from both numerator and denominator:

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

We can split this fraction:

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

Using the definitions \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\):

Expression \( = \tan \theta + \sec \theta\)

Now, let's look at the options provided.

Option 1 is \(\frac{1}{\sec \theta-\tan \theta}\). Recall the identity \(\sec^2 \theta - \tan^2 \theta = 1\). This can be factored as \((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\). Dividing both sides by \((\sec \theta - \tan \theta)\), we get \(\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta}\).

Our simplified expression is \(\sec \theta + \tan \theta\), which is equal to \(\frac{1}{\sec \theta - \tan \theta}\).

This matches Option 1.

Let's briefly check Option 2, \(\frac{1}{\operatorname{cosec} \theta-\cot \theta}\). Using the identity \(\operatorname{cosec}^2 \theta - \cot^2 \theta = 1\), we get \((\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1\), so \(\frac{1}{\operatorname{cosec} \theta - \cot \theta} = \operatorname{cosec} \theta + \cot \theta\). Our simplified expression \(\sec \theta + \tan \theta\) is not generally equal to \(\operatorname{cosec} \theta + \cot \theta\).

Thus, the given expression is equal to \(\frac{1}{\sec \theta-\tan \theta}\).