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QuestionProve that 1+sin⁡xcos⁡x+cos⁡x1+sin⁡x≡2sec⁡x\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} \equiv 2 \sec x.

Studdy Solution

STEP 1

Assumptions1. We are given the trigonometric equation 1+sin⁡xcos⁡x+cos⁡x1+sin⁡x≡sec⁡x\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} \equiv \sec x. . We need to prove the left side of the equation is equivalent to the right side.
3. We will use trigonometric identities and algebraic manipulation to solve this problem.

STEP 2

First, let's simplify the left side of the equation. We can do this by adding the two fractions. To add fractions, we need a common denominator. In this case, the common denominator is cos⁡x(1+sin⁡x)\cos x (1+\sin x).
1+sin⁡xcos⁡x+cos⁡x1+sin⁡x=(1+sin⁡x)2+cos⁡2xcos⁡x(1+sin⁡x)\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} = \frac{(1+\sin x)^2 + \cos^2 x}{\cos x (1+\sin x)}

STEP 3

Now, let's simplify the numerator of the fraction. We can do this by expanding (1+sin⁡x)2(1+\sin x)^2 and using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1.
(1+sin⁡x)2+cos⁡2xcos⁡x(1+sin⁡x)=1+2sin⁡x+sin⁡2x+cos⁡2xcos⁡x(1+sin⁡x)=2+2sin⁡xcos⁡x(1+sin⁡x)\frac{(1+\sin x)^2 + \cos^2 x}{\cos x (1+\sin x)} = \frac{1 +2\sin x + \sin^2 x + \cos^2 x}{\cos x (1+\sin x)} = \frac{2 +2\sin x}{\cos x (1+\sin x)}

STEP 4

Next, let's simplify the denominator of the fraction. We can do this by multiplying cos⁡x\cos x and (1+sin⁡x)(1+\sin x).
2+2sin⁡xcos⁡x(1+sin⁡x)=2+2sin⁡xcos⁡x+sin⁡xcos⁡x\frac{2 +2\sin x}{\cos x (1+\sin x)} = \frac{2 +2\sin x}{\cos x + \sin x \cos x}

STEP 5

Now, let's simplify the fraction by dividing the numerator and the denominator by 22.
2+2sin⁡xcos⁡x+sin⁡xcos⁡x=1+sin⁡xcos⁡x2+sin⁡xcos⁡x2\frac{2 +2\sin x}{\cos x + \sin x \cos x} = \frac{1 + \sin x}{\frac{\cos x}{2} + \frac{\sin x \cos x}{2}}

STEP 6

Next, let's simplify the equation by multiplying both sides by 22.
1+sin⁡xcos⁡x2+sin⁡xcos⁡x2=2sec⁡x\frac{1 + \sin x}{\frac{\cos x}{2} + \frac{\sin x \cos x}{2}} =2 \sec x21+sin⁡xcos⁡x+sin⁡xcos⁡x=2sec⁡x2 \frac{1 + \sin x}{\cos x + \sin x \cos x} =2 \sec x

STEP 7

Finally, let's simplify the right side of the equation. We can do this by using the identity sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.
21+sin⁡xcos⁡x+sin⁡xcos⁡x=21cos⁡x2 \frac{1 + \sin x}{\cos x + \sin x \cos x} =2 \frac{1}{\cos x}

STEP 8

Now, let's simplify the equation by cancelling out the 22 on both sides.
1+sin⁡xcos⁡x+sin⁡xcos⁡x=1cos⁡x\frac{1 + \sin x}{\cos x + \sin x \cos x} = \frac{1}{\cos x}Therefore, the left side of the equation is equivalent to the right side, and the original equation is proven to be true.

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