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Math  /  Algebra

QuestionUse identities to find an expression equivalent to 1+sin⁡(6x)1−sin⁡(6x)−1−sin⁡(6x)1+sin⁡(6x)\frac{1+\sin (6 x)}{1-\sin (6 x)}-\frac{1-\sin (6 x)}{1+\sin (6 x)}

Studdy Solution

STEP 1

1. We are asked to simplify the given expression using trigonometric identities.
2. The expression involves the sine function and can be simplified using algebraic manipulation and trigonometric identities.

STEP 2

1. Simplify the expression by combining the fractions.
2. Use trigonometric identities to further simplify the expression.

STEP 3

Combine the fractions by finding a common denominator. The common denominator for the two fractions is (1−sin⁡(6x))(1+sin⁡(6x))(1 - \sin(6x))(1 + \sin(6x)).
1+sin⁡(6x)1−sin⁡(6x)−1−sin⁡(6x)1+sin⁡(6x)=(1+sin⁡(6x))2−(1−sin⁡(6x))2(1−sin⁡(6x))(1+sin⁡(6x))\frac{1+\sin (6 x)}{1-\sin (6 x)} - \frac{1-\sin (6 x)}{1+\sin (6 x)} = \frac{(1+\sin(6x))^2 - (1-\sin(6x))^2}{(1-\sin(6x))(1+\sin(6x))}

STEP 4

Simplify the numerator using the difference of squares formula:
(1+sin⁡(6x))2−(1−sin⁡(6x))2=[(1+sin⁡(6x))−(1−sin⁡(6x))][(1+sin⁡(6x))+(1−sin⁡(6x))](1+\sin(6x))^2 - (1-\sin(6x))^2 = [(1+\sin(6x)) - (1-\sin(6x))][(1+\sin(6x)) + (1-\sin(6x))]
=[2sin⁡(6x)][2]=4sin⁡(6x)= [2\sin(6x)][2] = 4\sin(6x)

STEP 5

Simplify the denominator using the identity for the difference of squares:
(1−sin⁡(6x))(1+sin⁡(6x))=1−(sin⁡(6x))2=cos⁡2(6x)(1-\sin(6x))(1+\sin(6x)) = 1 - (\sin(6x))^2 = \cos^2(6x)

STEP 6

Substitute the simplified numerator and denominator back into the expression:
4sin⁡(6x)cos⁡2(6x)\frac{4\sin(6x)}{\cos^2(6x)}
This simplifies to:
4⋅sin⁡(6x)cos⁡2(6x)=4⋅tan⁡(6x)⋅sec⁡(6x)4 \cdot \frac{\sin(6x)}{\cos^2(6x)} = 4 \cdot \tan(6x) \cdot \sec(6x)
The equivalent expression is:
4tan⁡(6x)sec⁡(6x)\boxed{4 \tan(6x) \sec(6x)}

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