To introduce tan θ and sec θ into the expression, we divide every term in both the numerator and the denominator by cos θ. This is a common technique when dealing with trigonometric identities involving sin θ and cos θ to transform them into tan θ and sec θ.

To introduce tan θ and sec θ into the expression, we divide every term in both the numerator and the denominator by cos θ. This is a common technique when dealing with trigonometric identities involving sin θ and cos θ to transform them into tan θ and sec θ.

We factor the term (sec² θ - tan² θ) using the difference of squares formula, a² - b² = (a-b)(a+b). Then, we factor out the common term (tan θ + sec θ) from the numerator.

Notice that the term (1 - sec θ + tan θ) in the numerator is identical to the denominator (tan θ - sec θ + 1). We can cancel these terms, simplifying the expression to tan θ + sec θ.

Finally, we multiply the numerator and denominator by (sec θ - tan θ). Using the identity sec² θ - tan² θ = 1, the numerator simplifies to 1, which matches the Right Hand Side (RHS) of the original identity.

Finally, we multiply the numerator and denominator by (sec θ - tan θ). Using the identity sec² θ - tan² θ = 1, the numerator simplifies to 1, which matches the Right Hand Side (RHS) of the original identity.

Prove that : $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$ .

Answer: $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$

Step-by-step solution

Step 1: Start with the Left Hand Side (LHS)

We begin by considering the Left Hand Side (LHS) of the given identity. Our goal is to manipulate this expression algebraically until it matches the Right Hand Side (RHS).

Step 2: Divide numerator and denominator by $\cos \theta$

To introduce $\tan \theta$ and $\sec \theta$ into the expression, we divide every term in both the numerator and the denominator by $\cos \theta$ . This is a common technique when dealing with trigonometric identities involving $\sin \theta$ and $\cos \theta$ to transform them into $\tan \theta$ and $\sec \theta$ .

Step 3: Rearrange terms and use identity

We rearrange the terms in the numerator to group $\tan \theta + \sec \theta$ . Then, we replace the constant '1' in the numerator with the identity $\sec^2 \theta - \tan^2 \theta = 1$ . This step is crucial for factoring later.

Step 4: Factor the numerator

We factor the term $(\sec^2 \theta - \tan^2 \theta)$ using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$ . Then, we factor out the common term $(\tan \theta + \sec \theta)$ from the numerator.

Step 5: Simplify the expression

Notice that the term $(1 - \sec \theta + \tan \theta)$ in the numerator is identical to the denominator $(\tan \theta - \sec \theta + 1)$ . We can cancel these terms, simplifying the expression to $\tan \theta + \sec \theta$ .

Step 6: Convert to the form of RHS

Finally, we multiply the numerator and denominator by $(\sec \theta - \tan \theta)$ . Using the identity $\sec^2 \theta - \tan^2 \theta = 1$ , the numerator simplifies to 1, which matches the Right Hand Side (RHS) of the original identity.

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Prove that : sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θ\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}sinθ+cosθ−1sinθ−cosθ+1​=secθ−tanθ1​.