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4. Prove the following identities: \( 4.1 \quad \cos ^{2} x+\sin x \cdot \cos x \cdot \tan x=1 \) -sit \( 4.2 \quad \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=2 \cos ^{2} \theta-1 \)

Ask by Ortega Sandoval. in South Africa
Feb 21,2025

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Answer

Both identities are proven to be true.

Solution

Verify the identity by following steps: - step0: Verify: \(\cos^{2}\left(x\right)+\sin\left(x\right)\cos\left(x\right)\tan\left(x\right)=1\) - step1: Choose a side to work on: \(1=1\) - step2: Verify the identity: \(\textrm{true}\) Determine whether the expression \( (1 - \tan^2(\theta)) / (1 + \tan^2(\theta)) = 2 * \cos^2(\theta) - 1 \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\left(1-\tan^{2}\left(\theta \right)\right)}{\left(1+\tan^{2}\left(\theta \right)\right)}=2\cos^{2}\left(\theta \right)-1\) - step1: Choose a side to work on: \(\cos\left(2\theta \right)=2\cos^{2}\left(\theta \right)-1\) - step2: Choose the other side to work on: \(\cos\left(2\theta \right)=\cos\left(2\theta \right)\) - step3: Verify the identity: \(\textrm{true}\) Both identities have been proven to be true: 1. For the first identity: \[ \cos^2(x) + \sin(x) \cdot \cos(x) \cdot \tan(x) = 1 \] This identity is true. 2. For the second identity: \[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = 2 \cos^2(\theta) - 1 \] This identity is also true. Thus, both identities hold as valid mathematical statements.

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To tackle the first identity \( \cos^{2} x + \sin x \cdot \cos x \cdot \tan x = 1 \), let's rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \). Substituting this into the equation gives: \[ \cos^{2} x + \sin x \cdot \cos x \cdot \left(\frac{\sin x}{\cos x}\right) = \cos^{2} x + \sin^{2} x = 1 \] This uses the Pythagorean identity \( \sin^{2} x + \cos^{2} x = 1 \). Now for the identity \( \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = 2 \cos^{2} \theta - 1 \), we start by expressing \( \tan^{2} \theta \) as \( \frac{\sin^{2} \theta}{\cos^{2} \theta} \): \[ \frac{1 - \frac{\sin^{2} \theta}{\cos^{2} \theta}}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} = \frac{\cos^{2} \theta - \sin^{2} \theta}{\cos^{2} \theta + \sin^{2} \theta} = \cos^{2} \theta - \sin^{2} \theta \] Since \( \cos^{2} \theta + \sin^{2} \theta = 1 \), we can simplify further to show this equals \( 2 \cos^{2} \theta - 1 \). Both identities hold true!

Related Questions

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