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prove the following identities: \( \begin{array}{ll}\text { (1) } \cos x \sin x+\sin ^{2} x \tan x=\tan x & \text { (2) } \frac{\tan A \cdot \sin 2 A}{1-\cos A}=2+2 \cos A \\ \text { (3) } \frac{\cos \alpha+\sin \alpha}{\cos 2 \alpha}=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos \alpha-\sin \alpha} & \text { (4) } \frac{\sin 2 x-2 \sin x}{\cos 2 x-1}=\frac{1}{\sin x}-\frac{1}{\tan x} \\ \text { (5) } \frac{\sin \theta-\sin 2 \theta}{\cos \theta-\cos 2 \theta-1}=\tan \theta & \text { (6) } \frac{\cos 2 \alpha+\cos \alpha}{\sin 2 \alpha-\sin \alpha}=\frac{\cos \alpha+1}{\sin \alpha}\end{array} \)

Ask by Ortega Reed. in South Africa
Mar 12,2025

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**Summary of Identity Verifications:** - **Identity (1)**: True - **Identity (2)**: False - **Identity (3)**: Needs further verification - **Identity (4)**: Needs further verification - **Identity (5)**: True - **Identity (6)**: Needs further verification

Solución

Verify the identity by following steps: - step0: Verify: \(\frac{\left(\sin\left(\theta \right)-\sin\left(2\theta \right)\right)}{\left(\cos\left(\theta \right)-\cos\left(2\theta \right)-1\right)}=\tan\left(\theta \right)\) - step1: Choose a side to work on: \(\tan\left(\theta \right)=\tan\left(\theta \right)\) - step2: Verify the identity: \(\textrm{true}\) Determine whether the expression \( \cos(x)\sin(x) + \sin^2(x)\tan(x) = \tan(x) \) is always true. Verify the identity by following steps: - step0: Verify: \(\cos\left(x\right)\sin\left(x\right)+\sin^{2}\left(x\right)\tan\left(x\right)=\tan\left(x\right)\) - step1: Choose a side to work on: \(\tan\left(x\right)=\tan\left(x\right)\) - step2: Verify the identity: \(\textrm{true}\) Determine whether the expression \( (\tan(A) * \sin(2A))/(1 - \cos(A)) = 2 + 2\cos(A) \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\left(\tan\left(A\right)\sin\left(2A\right)\right)}{\left(1-\cos\left(A\right)\right)}=2+2\cos\left(A\right)\) - step1: Choose a side to work on: \(\frac{\tan\left(A\right)\sin\left(2A\right)}{1-\cos\left(A\right)}=2+2\cos\left(A\right)\) - step2: Verify the identity: \(\textrm{false}\) Let's analyze the identities one by one based on the results obtained. 1. **Identity (1)**: \[ \cos x \sin x + \sin^2 x \tan x = \tan x \] - Result: **True**. This identity holds. 2. **Identity (2)**: \[ \frac{\tan A \cdot \sin 2A}{1 - \cos A} = 2 + 2 \cos A \] - Result: **False**. This identity does not hold. 3. **Identity (3)**: \[ \frac{\cos \alpha + \sin \alpha}{\cos 2\alpha} = \frac{\cos^2 \alpha + \sin^2 \alpha}{\cos \alpha - \sin \alpha} \] - Result: **Calculation timeout**. This identity needs further verification. 4. **Identity (4)**: \[ \frac{\sin 2x - 2\sin x}{\cos 2x - 1} = \frac{1}{\sin x} - \frac{1}{\tan x} \] - Result: **Service not available**. This identity needs further verification. 5. **Identity (5)**: \[ \frac{\sin \theta - \sin 2\theta}{\cos \theta - \cos 2\theta - 1} = \tan \theta \] - Result: **True**. This identity holds. 6. **Identity (6)**: \[ \frac{\cos 2\alpha + \cos \alpha}{\sin 2\alpha - \sin \alpha} = \frac{\cos \alpha + 1}{\sin \alpha} \] - Result: **Calculation timeout**. This identity needs further verification. ### Summary of Results: - **True**: (1), (5) - **False**: (2) - **Timeout**: (3), (4), (6) For identities (3), (4), and (6), further analysis or a different approach may be required to verify their validity. If you would like to proceed with any specific identity or need further assistance, please let me know!

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Sure! Let's tackle two of the identities. 1. To prove \( \cos x \sin x + \sin^2 x \tan x = \tan x \): Start with the left side: \[ \cos x \sin x + \sin^2 x \tan x = \cos x \sin x + \sin^2 x \frac{\sin x}{\cos x} = \cos x \sin x + \frac{\sin^3 x}{\cos x} \] Combine the terms over a common denominator, which is \( \cos x \): \[ = \frac{\cos^2 x \sin x + \sin^3 x}{\cos x} = \frac{\sin x (\cos^2 x + \sin^2 x)}{\cos x} \] Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \): \[ = \frac{\sin x}{\cos x} = \tan x \] Hence, \( \cos x \sin x + \sin^2 x \tan x = \tan x \) is proven. 2. For the identity \( \frac{\tan A \cdot \sin 2A}{1 - \cos A} = 2 + 2 \cos A \): Start with the left side: \[ \frac{\tan A \cdot \sin 2A}{1 - \cos A} = \frac{\frac{\sin A}{\cos A} \cdot 2 \sin A \cos A}{1 - \cos A} = \frac{2 \sin^2 A}{\cos A (1 - \cos A)} \] We can express \( 1 - \cos A \) in terms of \( \sin^2 A \): \[ 1 - \cos A = \frac{\sin^2 A}{1 + \cos A} \] Therefore: \[ \frac{2 \sin^2 A}{\cos A \cdot \frac{\sin^2 A}{1 + \cos A}} = \frac{2(1 + \cos A)}{\cos A} \] Simplifying gives: \[ = 2 + 2 \cos A \] Thus, the identity is proven.

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