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Math  /  Trigonometry

Questiona cos⁡α1−cos⁡2α1−sin⁡2α\frac{\cos \alpha \sqrt{1-\cos ^{2} \alpha}}{1-\sin ^{2} \alpha}

Studdy Solution

STEP 1

1. We are asked to simplify the given trigonometric expression.
2. We will use trigonometric identities to simplify the expression.

STEP 2

1. Simplify the numerator of the expression.
2. Simplify the denominator of the expression.
3. Simplify the entire expression using the results from the previous steps.

STEP 3

Simplify the numerator cos⁡α1−cos⁡2α\cos \alpha \sqrt{1-\cos ^{2} \alpha}.
Using the Pythagorean identity, we know that:
1−cos⁡2α=sin⁡2α 1 - \cos^2 \alpha = \sin^2 \alpha
Thus, the numerator becomes:
cos⁡α⋅sin⁡2α=cos⁡α⋅∣sin⁡α∣ \cos \alpha \cdot \sqrt{\sin^2 \alpha} = \cos \alpha \cdot |\sin \alpha|
Assuming α\alpha is in a range where sin⁡α\sin \alpha is non-negative, we have:
cos⁡α⋅sin⁡α \cos \alpha \cdot \sin \alpha

STEP 4

Simplify the denominator 1−sin⁡2α1-\sin ^{2} \alpha.
Using the Pythagorean identity again, we have:
1−sin⁡2α=cos⁡2α 1 - \sin^2 \alpha = \cos^2 \alpha

STEP 5

Combine the simplified numerator and denominator:
The expression becomes:
cos⁡α⋅sin⁡αcos⁡2α \frac{\cos \alpha \cdot \sin \alpha}{\cos^2 \alpha}
Simplify by canceling a common factor of cos⁡α\cos \alpha:
sin⁡αcos⁡α \frac{\sin \alpha}{\cos \alpha}
This simplifies to:
tan⁡α \tan \alpha
The simplified expression is:
tan⁡α \boxed{\tan \alpha}

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