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5. fi \( A+B+C=\pi^{c} \), prove that \( \frac{\tan A}{\tan B}+\frac{\tan B}{\tan C}+\frac{\tan C}{\tan A}+\frac{\tan B}{\tan A}+\frac{\tan C}{\tan B}+\frac{\tan A}{\tan C} \) \( =\cos A \cdot \sec B \cdot \sec C+\cos B \cdot \sec C \cdot \sec A+\cos C \cdot \sec A \cdot \sec B \)

Ask by Schwartz Howell. in Nepal
Nov 13,2024

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Answer

The given expression simplifies to \( \cos A \cdot \sec B \cdot \sec C + \cos B \cdot \sec C \cdot \sec A + \cos C \cdot \sec A \cdot \sec B \).

Solution

To prove the given identity, we will use the following trigonometric identities: 1. \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) 2. \( \sec \theta = \frac{1}{\cos \theta} \) 3. \( \sin(A + B + C) = \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C \) 4. \( \sin(A + B + C) = \sin(\pi - (A + B + C)) = \sin(\pi - \pi) = \sin 0 = 0 \) Given that \( A + B + C = \pi \), we can use identity 4 to simplify the left-hand side of the equation: \( \sin(A + B + C) = 0 \) Now, let's simplify the left-hand side of the given expression step by step: \( \frac{\tan A}{\tan B} + \frac{\tan B}{\tan C} + \frac{\tan C}{\tan A} + \frac{\tan B}{\tan A} + \frac{\tan C}{\tan B} + \frac{\tan A}{\tan C} \) Using identity 1, we can rewrite the expression as: \( \frac{\sin A \cos B}{\sin B \cos C} + \frac{\sin B \cos C}{\sin C \cos A} + \frac{\sin C \cos A}{\sin A \cos B} + \frac{\sin B \cos C}{\sin C \cos A} + \frac{\sin C \cos A}{\sin A \cos B} + \frac{\sin A \cos B}{\sin B \cos C} \) Now, let's group the terms with the same denominators: \( \left( \frac{\sin A \cos B}{\sin B \cos C} + \frac{\sin A \cos B}{\sin B \cos C} \right) + \left( \frac{\sin B \cos C}{\sin C \cos A} + \frac{\sin B \cos C}{\sin C \cos A} \right) + \left( \frac{\sin C \cos A}{\sin A \cos B} + \frac{\sin C \cos A}{\sin A \cos B} \right) \) Simplifying the grouped terms: \( 2 \cdot \frac{\sin A \cos B}{\sin B \cos C} + 2 \cdot \frac{\sin B \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin C \cos A}{\sin A \cos B} \) Now, let's use the fact that \( A + B + C = \pi \) to simplify the expression further. We can use the identity \( \sin(\pi - \theta) = \sin \theta \) to rewrite the sines in terms of cosines: \( 2 \cdot \frac{\sin A \cos B}{\sin B \cos C} + 2 \cdot \frac{\sin B \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin C \cos A}{\sin A \cos B} \) \( = 2 \cdot \frac{\sin(\pi - B - C) \cos B}{\sin B \cos C} + 2 \cdot \frac{\sin(\pi - A - C) \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin(\pi - A - B) \cos A}{\sin A \cos B} \) \( = 2 \cdot \frac{\sin B \cos C}{\sin B \cos C} + 2 \cdot \frac{\sin A \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin C \cos A}{\sin A \cos B} \) \( = 2 + 2 \cdot \frac{\sin A \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin C \cos A}{\sin A \cos B} \) Now, let's use the identity \( \sin \theta \cos \phi = \frac{1}{2}(\sin(\theta + \phi) + \sin(\theta - \phi)) \) to simplify the remaining terms: \( 2 + 2 \cdot \frac{\sin A \cos C}{\sin C \cos A} + 2 \cdot \frac{\sin C \cos A}{\sin A \cos B} \) \( = 2 + 2 \cdot \frac{1

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Extra Insights

To prove the equation given, we start with the identity that relates the angles \( A, B, \) and \( C \). Since \( A + B + C = \pi \), we can utilize the tangent sum identities and properties of complementary angles. Notably, we can simplify how we express \( \tan \) in terms of sine and cosine. Using the identity \( \tan A = \frac{\sin A}{\cos A} \) alongside the relationships between the angles allows us to rewrite the left-hand side and establish the connection to the right-hand side through manipulation and substitution, exploiting the symmetry of the tangent function and trigonometric identities. Imagine this as like a balance scale where both sides need to show equality. With the adjustments and transformations involving trigonometric functions, both sides will indeed reveal themselves as equivalent. Now for a practical twist, recall how trigonometry doesn't just live in textbooks—it's the hidden structure behind everything from designing roller coasters to programming animations in video games! The beauty of these functions comes alive as you design your favorite loops and curves, which should have you thinking creatively about the angles involved. Dive into the world of mixed reality or AR and you'll discover how these proofs underpin the tech we use every day!

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