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Sagot :
Sure, let's write the given double-angled trigonometric expression in terms of [tex]\(\tan \theta\)[/tex].
We start with the expression:
[tex]\[ \frac{\tan 2\theta}{\cos 2\theta}(1 - \sin 2\theta) \][/tex]
First, recall the double angle formulas for [tex]\(\tan\)[/tex], [tex]\(\cos\)[/tex], and [tex]\(\sin\)[/tex]:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
[tex]\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \][/tex]
[tex]\[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Substitute these into the given expression:
[tex]\[ \frac{\tan 2\theta}{\cos 2\theta}(1 - \sin 2\theta) \][/tex]
[tex]\[ = \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} \left(1 - \frac{2 \tan \theta}{1 + \tan^2 \theta}\right) \][/tex]
Simplify the fraction part first:
[tex]\[ \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} = \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = \frac{2 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2} \][/tex]
Next, simplify the term inside the parentheses:
[tex]\[ 1 - \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{1 + \tan^2 \theta - 2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Now, multiply the simplified parts together:
[tex]\[ \frac{2 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2} \cdot \frac{1 + \tan^2 \theta - 2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Combine the fractions:
[tex]\[ \frac{2 \tan \theta (1 + \tan^2 \theta)(1 + \tan^2 \theta - 2 \tan \theta)}{(1 - \tan^2 \theta)^2 (1 + \tan^2 \theta)} \][/tex]
Simplify the numerator:
[tex]\[ 2 \tan \theta (1 + \tan^2 \theta - 2 \tan \theta) = 2 \tan \theta \left((1 + \tan^2 \theta) (1 + \tan^2 \theta) -2 \tan \theta \right) \][/tex]
Thus, the final simplified expression is:
[tex]\[ \boxed{2\left(1 - \frac{2 \tan \theta}{\tan^2 \theta + 1}\right) \left( \tan^2 \theta + 1 \right) \frac{\tan \theta}{(1 - \tan^2 \theta)^2}} \][/tex]
This is how you express [tex]\(\frac{\tan 2 \theta}{\cos 2 \theta}(1-\sin 2 \theta)\)[/tex] in terms of [tex]\(\tan \theta\)[/tex].
We start with the expression:
[tex]\[ \frac{\tan 2\theta}{\cos 2\theta}(1 - \sin 2\theta) \][/tex]
First, recall the double angle formulas for [tex]\(\tan\)[/tex], [tex]\(\cos\)[/tex], and [tex]\(\sin\)[/tex]:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
[tex]\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \][/tex]
[tex]\[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Substitute these into the given expression:
[tex]\[ \frac{\tan 2\theta}{\cos 2\theta}(1 - \sin 2\theta) \][/tex]
[tex]\[ = \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} \left(1 - \frac{2 \tan \theta}{1 + \tan^2 \theta}\right) \][/tex]
Simplify the fraction part first:
[tex]\[ \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}} = \frac{2 \tan \theta}{1 - \tan^2 \theta} \cdot \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = \frac{2 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2} \][/tex]
Next, simplify the term inside the parentheses:
[tex]\[ 1 - \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{1 + \tan^2 \theta - 2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Now, multiply the simplified parts together:
[tex]\[ \frac{2 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2} \cdot \frac{1 + \tan^2 \theta - 2 \tan \theta}{1 + \tan^2 \theta} \][/tex]
Combine the fractions:
[tex]\[ \frac{2 \tan \theta (1 + \tan^2 \theta)(1 + \tan^2 \theta - 2 \tan \theta)}{(1 - \tan^2 \theta)^2 (1 + \tan^2 \theta)} \][/tex]
Simplify the numerator:
[tex]\[ 2 \tan \theta (1 + \tan^2 \theta - 2 \tan \theta) = 2 \tan \theta \left((1 + \tan^2 \theta) (1 + \tan^2 \theta) -2 \tan \theta \right) \][/tex]
Thus, the final simplified expression is:
[tex]\[ \boxed{2\left(1 - \frac{2 \tan \theta}{\tan^2 \theta + 1}\right) \left( \tan^2 \theta + 1 \right) \frac{\tan \theta}{(1 - \tan^2 \theta)^2}} \][/tex]
This is how you express [tex]\(\frac{\tan 2 \theta}{\cos 2 \theta}(1-\sin 2 \theta)\)[/tex] in terms of [tex]\(\tan \theta\)[/tex].
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