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].
