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Sagot :
Sure! Let's solve the given trigonometric equation step-by-step.
The equation we need to simplify is:
[tex]\[ \cot ^2 \theta \sin ^2 \theta + \tan ^2 \theta \cos ^2 \theta = 1 \][/tex]
First, let's rewrite the trigonometric identities in terms of sine and cosine. We know that:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Thus:
[tex]\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
[tex]\[ \tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Substituting these into the original equation, we get:
[tex]\[ \left(\frac{\cos^2 \theta}{\sin^2 \theta}\right) \sin^2 \theta + \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos^2 \theta \][/tex]
Simplify each term separately:
[tex]\[ \frac{\cos^2 \theta}{\sin^2 \theta} \cdot \sin^2 \theta = \cos^2 \theta \][/tex]
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \cos^2 \theta = \sin^2 \theta \][/tex]
Therefore, the equation simplifies to:
[tex]\[ \cos^2 \theta + \sin^2 \theta \][/tex]
We use the Pythagorean identity for sine and cosine:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Hence, we have shown that:
[tex]\[ \cot ^2 \theta \sin ^2 \theta + \tan ^2 \theta \cos ^2 \theta = 1 \][/tex]
So the given equation is indeed equal to 1.
The equation we need to simplify is:
[tex]\[ \cot ^2 \theta \sin ^2 \theta + \tan ^2 \theta \cos ^2 \theta = 1 \][/tex]
First, let's rewrite the trigonometric identities in terms of sine and cosine. We know that:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Thus:
[tex]\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
[tex]\[ \tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta} \][/tex]
Substituting these into the original equation, we get:
[tex]\[ \left(\frac{\cos^2 \theta}{\sin^2 \theta}\right) \sin^2 \theta + \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos^2 \theta \][/tex]
Simplify each term separately:
[tex]\[ \frac{\cos^2 \theta}{\sin^2 \theta} \cdot \sin^2 \theta = \cos^2 \theta \][/tex]
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \cos^2 \theta = \sin^2 \theta \][/tex]
Therefore, the equation simplifies to:
[tex]\[ \cos^2 \theta + \sin^2 \theta \][/tex]
We use the Pythagorean identity for sine and cosine:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Hence, we have shown that:
[tex]\[ \cot ^2 \theta \sin ^2 \theta + \tan ^2 \theta \cos ^2 \theta = 1 \][/tex]
So the given equation is indeed equal to 1.
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