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Sagot :
Let's begin with the given equation:
[tex]\[ \cos^4 \theta + \cos^2 \theta = 1 \][/tex]
To demonstrate that [tex]\(\tan^4 \theta + \tan^2 \theta = 1\)[/tex], we start by expressing [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Next, let's square [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left( \frac{\sin \theta}{\cos \theta} \right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Now, let's square [tex]\(\tan^2 \theta\)[/tex] to find [tex]\(\tan^4 \theta\)[/tex]:
[tex]\[ \tan^4 \theta = \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)^2 = \frac{\sin^4 \theta}{\cos^4 \theta} \][/tex]
We now need to show that:
[tex]\[ \frac{\sin^4 \theta}{\cos^4 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = 1 \][/tex]
Let's re-express our terms:
[tex]\[ \tan^4 \theta + \tan^2 \theta = \frac{\sin^4 \theta}{\cos^4 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
We will use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
We can express [tex]\(\sin^2 \theta\)[/tex] as:
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
We square this expression to find [tex]\(\sin^4 \theta\)[/tex]:
[tex]\[ \sin^4 \theta = (1 - \cos^2 \theta)^2 = 1 - 2\cos^2 \theta + \cos^4 \theta \][/tex]
Now substitute [tex]\(\sin^4 \theta\)[/tex] and [tex]\(\sin^2 \theta\)[/tex] into our equation in terms of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \frac{1 - 2\cos^2 \theta + \cos^4 \theta}{\cos^4 \theta} + \frac{1 - \cos^2 \theta}{\cos^2 \theta} \][/tex]
Let's simplify each term separately:
[tex]\[ \frac{1 - 2\cos^2 \theta + \cos^4 \theta}{\cos^4 \theta} = \frac{1}{\cos^4 \theta} - \frac{2\cos^2 \theta}{\cos^4 \theta} + \frac{\cos^4 \theta}{\cos^4 \theta} = \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + 1 \][/tex]
[tex]\[ \frac{1 - \cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} - 1 \][/tex]
Putting it all together:
[tex]\[ \left( \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + 1 \right) + \left( \frac{1}{\cos^2 \theta} - 1 \right) \][/tex]
Combine like terms:
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} + 1 - 1 \][/tex]
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{1}{\cos^2 \theta} \][/tex]
Recall the original equation:
[tex]\[ \cos^4 \theta + \cos^2 \theta = 1 \][/tex]
By multiplying both sides of this equation by [tex]\(\frac{1}{\cos^4 \theta}\)[/tex]:
[tex]\[ 1 + \frac{1}{\cos^2 \theta} = \frac{1}{\cos^4 \theta} \][/tex]
So, our simplified expression becomes:
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{1}{\cos^2 \theta} = 1 \][/tex]
Thus proving:
[tex]\[ \tan^4 \theta + \tan^2 \theta = 1 \][/tex]
[tex]\[ \cos^4 \theta + \cos^2 \theta = 1 \][/tex]
To demonstrate that [tex]\(\tan^4 \theta + \tan^2 \theta = 1\)[/tex], we start by expressing [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Next, let's square [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left( \frac{\sin \theta}{\cos \theta} \right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Now, let's square [tex]\(\tan^2 \theta\)[/tex] to find [tex]\(\tan^4 \theta\)[/tex]:
[tex]\[ \tan^4 \theta = \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)^2 = \frac{\sin^4 \theta}{\cos^4 \theta} \][/tex]
We now need to show that:
[tex]\[ \frac{\sin^4 \theta}{\cos^4 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = 1 \][/tex]
Let's re-express our terms:
[tex]\[ \tan^4 \theta + \tan^2 \theta = \frac{\sin^4 \theta}{\cos^4 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
We will use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
We can express [tex]\(\sin^2 \theta\)[/tex] as:
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
We square this expression to find [tex]\(\sin^4 \theta\)[/tex]:
[tex]\[ \sin^4 \theta = (1 - \cos^2 \theta)^2 = 1 - 2\cos^2 \theta + \cos^4 \theta \][/tex]
Now substitute [tex]\(\sin^4 \theta\)[/tex] and [tex]\(\sin^2 \theta\)[/tex] into our equation in terms of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \frac{1 - 2\cos^2 \theta + \cos^4 \theta}{\cos^4 \theta} + \frac{1 - \cos^2 \theta}{\cos^2 \theta} \][/tex]
Let's simplify each term separately:
[tex]\[ \frac{1 - 2\cos^2 \theta + \cos^4 \theta}{\cos^4 \theta} = \frac{1}{\cos^4 \theta} - \frac{2\cos^2 \theta}{\cos^4 \theta} + \frac{\cos^4 \theta}{\cos^4 \theta} = \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + 1 \][/tex]
[tex]\[ \frac{1 - \cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} - 1 \][/tex]
Putting it all together:
[tex]\[ \left( \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + 1 \right) + \left( \frac{1}{\cos^2 \theta} - 1 \right) \][/tex]
Combine like terms:
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{2}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} + 1 - 1 \][/tex]
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{1}{\cos^2 \theta} \][/tex]
Recall the original equation:
[tex]\[ \cos^4 \theta + \cos^2 \theta = 1 \][/tex]
By multiplying both sides of this equation by [tex]\(\frac{1}{\cos^4 \theta}\)[/tex]:
[tex]\[ 1 + \frac{1}{\cos^2 \theta} = \frac{1}{\cos^4 \theta} \][/tex]
So, our simplified expression becomes:
[tex]\[ \frac{1}{\cos^4 \theta} - \frac{1}{\cos^2 \theta} = 1 \][/tex]
Thus proving:
[tex]\[ \tan^4 \theta + \tan^2 \theta = 1 \][/tex]
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