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(4) hiven \( \frac{\sin \theta}{1+\cos \theta} \) 4.1 Proue that \( \frac{\sin \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\sin \theta} \) +2 For which value of \( \theta \) in the intoriul \( \theta \in(0 ; 360) \) is \( \frac{\sin \theta}{14 \cos \theta} \) andetingd

Ask by Chadwick Ball. in South Africa
Mar 13,2025

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To prove the identity, cross-multiply to get \( \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \). Simplifying the right side using the difference of squares gives \( 1 - \cos^2 \theta \), which equals \( \sin^2 \theta \) by the Pythagorean identity. Therefore, the identity holds true. For the second part, the expression \( \frac{\sin \theta}{14 \cos \theta} \) is undefined when the denominator is zero, i.e., when \( \cos \theta = 0 \). This occurs at \( \theta = 90^\circ \) and \( \theta = 270^\circ \) within the interval \( (0, 360) \).

Solution

Let's solve the problem step by step. ### Step 1: Prove that \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \) To prove this identity, we will start by cross-multiplying both sides: \[ \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \] The right-hand side can be simplified using the difference of squares: \[ (1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta \] Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute \( 1 - \cos^2 \theta \) with \( \sin^2 \theta \): \[ \sin^2 \theta = \sin^2 \theta \] This shows that both sides are equal, thus proving the identity: \[ \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \] ### Step 2: Find the value of \( \theta \) in the interval \( (0, 360) \) for which \( \frac{\sin \theta}{14 \cos \theta} \) is undefined. The expression \( \frac{\sin \theta}{14 \cos \theta} \) is undefined when the denominator is zero, which occurs when: \[ 14 \cos \theta = 0 \] This simplifies to: \[ \cos \theta = 0 \] The cosine function is zero at the following angles in the interval \( (0, 360) \): \[ \theta = 90^\circ \quad \text{and} \quad \theta = 270^\circ \] ### Final Answer 1. The identity \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \) is proven. 2. The values of \( \theta \) in the interval \( (0, 360) \) for which \( \frac{\sin \theta}{14 \cos \theta} \) is undefined are \( 90^\circ \) and \( 270^\circ \).

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To prove that \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \), we can cross-multiply to get \( \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \). The right side simplifies to \( 1 - \cos^2 \theta \), which equals \( \sin^2 \theta \) based on the Pythagorean Identity. Thus, both sides are indeed equivalent, validating the identity! Now, regarding the value of \( \theta \) for which \( \frac{\sin \theta}{14 \cos \theta} \) is undefined, this expression becomes undefined when the denominator equals zero, i.e., when \( \cos \theta = 0 \). This occurs at \( \theta = 90^\circ \) and \( \theta = 270^\circ \) within the interval \( (0, 360) \).

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