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Math  /  Trigonometry

Questionsin⁡xcos⁡x+1+cos⁡x+1sin⁡x=2esc⁡x\frac{\sin x}{\cos x+1}+\frac{\cos x+1}{\sin x}=2 \operatorname{esc} x

Studdy Solution

STEP 1

1. The equation involves trigonometric functions: sine, cosine, and cosecant.
2. We need to simplify and verify the given equation.
3. The identity csc⁡x=1sin⁡x \csc x = \frac{1}{\sin x} will be useful.
4. The equation should hold for values of x x where the trigonometric functions are defined.

STEP 2

1. Simplify the left-hand side of the equation.
2. Simplify the right-hand side of the equation.
3. Equate the simplified expressions and verify the identity.

STEP 3

Simplify the left-hand side of the equation:
sin⁡xcos⁡x+1+cos⁡x+1sin⁡x\frac{\sin x}{\cos x + 1} + \frac{\cos x + 1}{\sin x}
To combine these fractions, find a common denominator, which is sin⁡x(cos⁡x+1) \sin x (\cos x + 1) .

STEP 4

Combine the fractions using the common denominator:
sin⁡2x+(cos⁡x+1)2sin⁡x(cos⁡x+1)\frac{\sin^2 x + (\cos x + 1)^2}{\sin x (\cos x + 1)}
Expand the numerator:
sin⁡2x+cos⁡2x+2cos⁡x+1\sin^2 x + \cos^2 x + 2\cos x + 1
Using the Pythagorean identity sin⁡2x+cos⁡2x=1 \sin^2 x + \cos^2 x = 1 , the numerator becomes:
1+2cos⁡x+1=2+2cos⁡x1 + 2\cos x + 1 = 2 + 2\cos x
Thus, the left-hand side simplifies to:
2(1+cos⁡x)sin⁡x(cos⁡x+1)\frac{2(1 + \cos x)}{\sin x (\cos x + 1)}

STEP 5

Simplify the right-hand side of the equation:
2csc⁡x=2sin⁡x2 \csc x = \frac{2}{\sin x}

STEP 6

Equate the simplified expressions from both sides:
2(1+cos⁡x)sin⁡x(cos⁡x+1)=2sin⁡x\frac{2(1 + \cos x)}{\sin x (\cos x + 1)} = \frac{2}{\sin x}
Cancel 2 2 and sin⁡x \sin x from both sides:
1+cos⁡xcos⁡x+1=1\frac{1 + \cos x}{\cos x + 1} = 1
Since 1+cos⁡x=cos⁡x+1 1 + \cos x = \cos x + 1 , the equation holds true.
The given equation is verified as true for values of x x where the trigonometric functions are defined and the denominators are non-zero.

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