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Sagot :
Sure, let's solve the equation step-by-step to prove that [tex]\(\sec \theta = \pm \frac{\sqrt{17}}{4}\)[/tex].
We start with the equation:
[tex]\[ \frac{\cos \theta}{\tan \theta \cdot \sin \theta} = 16 \][/tex]
First, we'll simplify the left-hand side 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]
Now substitute [tex]\(\tan \theta\)[/tex] back into the original equation:
[tex]\[ \frac{\cos \theta}{\left(\frac{\sin \theta}{\cos \theta}\right) \cdot \sin \theta} = 16 \][/tex]
Simplify the denominator:
[tex]\[ \frac{\cos \theta}{\frac{\sin \theta^2}{\cos \theta}} = 16 \][/tex]
When we divide by a fraction, we multiply by its reciprocal:
[tex]\[ \frac{\cos \theta \cdot \cos \theta}{\sin \theta^2} = 16 \][/tex]
[tex]\[ \frac{\cos^2 \theta}{\sin^2 \theta} = 16 \][/tex]
This can be expressed as:
[tex]\[ \cot^2 \theta = 16 \][/tex]
Now, solving for [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \pm 4 \][/tex]
We know that:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
So:
[tex]\[ \frac{\cos \theta}{\sin \theta} = \pm 4 \][/tex]
Cross-multiplying:
[tex]\[ \cos \theta = \pm 4 \sin \theta \][/tex]
Next, let's find an expression for [tex]\(\sec \theta\)[/tex]. Recall that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]. Using the Pythagorean identity, [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
First, let's express [tex]\(\sin \theta\)[/tex] in terms of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
Substitute [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex] into [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ \sin^2 \theta + (4 \sin \theta)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + 16 \sin^2 \theta = 1 \][/tex]
[tex]\[ 17 \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = \frac{1}{17} \][/tex]
Thus:
[tex]\[ \sin \theta = \pm \frac{1}{\sqrt{17}} \][/tex]
Now substitute [tex]\(\sin \theta = \pm \frac{1}{\sqrt{17}}\)[/tex] back into [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex]:
[tex]\[ \cos \theta = \pm 4 \left(\pm \frac{1}{\sqrt{17}}\right) \][/tex]
[tex]\[ \cos \theta = \pm \frac{4}{\sqrt{17}} \][/tex]
Finally, we find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\pm \frac{4}{\sqrt{17}}} = \pm \frac{\sqrt{17}}{4} \][/tex]
Hence, we have proven that:
[tex]\[ \sec \theta = \pm \frac{\sqrt{17}}{4} \][/tex]
We start with the equation:
[tex]\[ \frac{\cos \theta}{\tan \theta \cdot \sin \theta} = 16 \][/tex]
First, we'll simplify the left-hand side 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]
Now substitute [tex]\(\tan \theta\)[/tex] back into the original equation:
[tex]\[ \frac{\cos \theta}{\left(\frac{\sin \theta}{\cos \theta}\right) \cdot \sin \theta} = 16 \][/tex]
Simplify the denominator:
[tex]\[ \frac{\cos \theta}{\frac{\sin \theta^2}{\cos \theta}} = 16 \][/tex]
When we divide by a fraction, we multiply by its reciprocal:
[tex]\[ \frac{\cos \theta \cdot \cos \theta}{\sin \theta^2} = 16 \][/tex]
[tex]\[ \frac{\cos^2 \theta}{\sin^2 \theta} = 16 \][/tex]
This can be expressed as:
[tex]\[ \cot^2 \theta = 16 \][/tex]
Now, solving for [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \pm 4 \][/tex]
We know that:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
So:
[tex]\[ \frac{\cos \theta}{\sin \theta} = \pm 4 \][/tex]
Cross-multiplying:
[tex]\[ \cos \theta = \pm 4 \sin \theta \][/tex]
Next, let's find an expression for [tex]\(\sec \theta\)[/tex]. Recall that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]. Using the Pythagorean identity, [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
First, let's express [tex]\(\sin \theta\)[/tex] in terms of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
Substitute [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex] into [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[ \sin^2 \theta + (4 \sin \theta)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + 16 \sin^2 \theta = 1 \][/tex]
[tex]\[ 17 \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = \frac{1}{17} \][/tex]
Thus:
[tex]\[ \sin \theta = \pm \frac{1}{\sqrt{17}} \][/tex]
Now substitute [tex]\(\sin \theta = \pm \frac{1}{\sqrt{17}}\)[/tex] back into [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex]:
[tex]\[ \cos \theta = \pm 4 \left(\pm \frac{1}{\sqrt{17}}\right) \][/tex]
[tex]\[ \cos \theta = \pm \frac{4}{\sqrt{17}} \][/tex]
Finally, we find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\pm \frac{4}{\sqrt{17}}} = \pm \frac{\sqrt{17}}{4} \][/tex]
Hence, we have proven that:
[tex]\[ \sec \theta = \pm \frac{\sqrt{17}}{4} \][/tex]
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