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Sagot :
Certainly! Let's solve the equation step by step.
Given the equation:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
First, recall the fundamental trigonometric identities:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Using these identities, we can rewrite the left-hand side of the equation:
[tex]\[ \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \][/tex]
Since they have a common denominator, we can combine these fractions:
[tex]\[ \sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta} \][/tex]
Now, substitute this back into the left-hand side of the original equation:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{1}{\frac{1 + \sin \theta}{\cos \theta}} \][/tex]
Inverting the fraction, we get:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]
So, the left-hand side of our original equation becomes:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]
Now, the original equation looks like this:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
For these two fractions to be equal, the numerators and the denominators multiplied crosswise should be equal:
[tex]\[ (\cos \theta)^2 = (1 + \sin \theta)(1 - \sin \theta) \][/tex]
Expanding the right-hand side using the difference of squares formula:
[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - (\sin \theta)^2 \][/tex]
Thus, we have:
[tex]\[ (\cos \theta)^2 = 1 - (\sin \theta)^2 \][/tex]
We know from the Pythagorean identity that:
[tex]\[ (\cos \theta)^2 + (\sin \theta)^2 = 1 \][/tex]
Therefore, substituting [tex]\(1 - (\sin \theta)^2\)[/tex] directly from the identity:
[tex]\[ (\cos \theta)^2 = (\cos \theta)^2 \][/tex]
The equation [tex]\(\frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta}\)[/tex] seems to hold when calculated based on this simplification. However, based on our observed mathematical result, we can deduce:
The equality [tex]\(\frac{1}{\sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta}\)[/tex] does not hold true for all values of [tex]\(\theta\)[/tex]. In other words, there is no general solution such that the left-hand side equals the right-hand side.
Given the equation:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
First, recall the fundamental trigonometric identities:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Using these identities, we can rewrite the left-hand side of the equation:
[tex]\[ \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \][/tex]
Since they have a common denominator, we can combine these fractions:
[tex]\[ \sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta} \][/tex]
Now, substitute this back into the left-hand side of the original equation:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{1}{\frac{1 + \sin \theta}{\cos \theta}} \][/tex]
Inverting the fraction, we get:
[tex]\[ \frac{1}{\sec \theta + \tan \theta} = \frac{\cos \theta}{1 + \sin \theta} \][/tex]
So, the left-hand side of our original equation becomes:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} \][/tex]
Now, the original equation looks like this:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta} \][/tex]
For these two fractions to be equal, the numerators and the denominators multiplied crosswise should be equal:
[tex]\[ (\cos \theta)^2 = (1 + \sin \theta)(1 - \sin \theta) \][/tex]
Expanding the right-hand side using the difference of squares formula:
[tex]\[ (1 + \sin \theta)(1 - \sin \theta) = 1 - (\sin \theta)^2 \][/tex]
Thus, we have:
[tex]\[ (\cos \theta)^2 = 1 - (\sin \theta)^2 \][/tex]
We know from the Pythagorean identity that:
[tex]\[ (\cos \theta)^2 + (\sin \theta)^2 = 1 \][/tex]
Therefore, substituting [tex]\(1 - (\sin \theta)^2\)[/tex] directly from the identity:
[tex]\[ (\cos \theta)^2 = (\cos \theta)^2 \][/tex]
The equation [tex]\(\frac{\cos \theta}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos \theta}\)[/tex] seems to hold when calculated based on this simplification. However, based on our observed mathematical result, we can deduce:
The equality [tex]\(\frac{1}{\sec \theta + \tan \theta} = \frac{1 - \sin \theta}{\cos \theta}\)[/tex] does not hold true for all values of [tex]\(\theta\)[/tex]. In other words, there is no general solution such that the left-hand side equals the right-hand side.
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