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(sin x)/(1-cos x)+(1-cos x)/(sin x)=2csc x

sin⁡x1−cos⁡x+1−cos⁡xsin⁡x=2csc⁡x \frac{\sin x}{1-\cos x}+\frac{1-\cos x}{\sin x}=2 \csc x

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Full solution

Q. sin⁡x1−cos⁡x+1−cos⁡xsin⁡x=2csc⁡x \frac{\sin x}{1-\cos x}+\frac{1-\cos x}{\sin x}=2 \csc x
  1. Combine fractions: Combine the two fractions by finding a common denominator, which is (sin⁡x)(1−cos⁡x)(\sin x)(1-\cos x).(sin⁡x)/(1−cos⁡x)⋅(sin⁡x)/(sin⁡x)+(1−cos⁡x)/(sin⁡x)⋅(1−cos⁡x)/(1−cos⁡x)=(sin⁡2x+(1−cos⁡x)2)/(sin⁡x)(1−cos⁡x)(\sin x)/(1-\cos x) \cdot (\sin x)/(\sin x) + (1-\cos x)/(\sin x) \cdot (1-\cos x)/(1-\cos x) = (\sin^2 x + (1-\cos x)^2)/(\sin x)(1-\cos x)
  2. Expand numerator: Expand the numerator: (sin⁡2x+(1−cos⁡x)2)(\sin^2 x + (1-\cos x)^2) becomes (sin⁡2x+1−2cos⁡x+cos⁡2x)(\sin^2 x + 1 - 2\cos x + \cos^2 x).\newline(sin⁡2x+1−2cos⁡x+cos⁡2x)/(sin⁡x)(1−cos⁡x)(\sin^2 x + 1 - 2\cos x + \cos^2 x)/(\sin x)(1-\cos x)
  3. Use Pythagorean identity: Use the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.\newline(1+1−2cos⁡x)/(sin⁡x)(1−cos⁡x)(1 + 1 - 2\cos x)/(\sin x)(1-\cos x)
  4. Simplify numerator: Simplify the numerator: 1+1−2cos⁡x1 + 1 - 2\cos x becomes 2−2cos⁡x2 - 2\cos x.\newline(2−2cos⁡x)/(sin⁡x)(1−cos⁡x)(2 - 2\cos x)/(\sin x)(1-\cos x)
  5. Factor out 22: Factor out the 22 in the numerator.\newline2(1−cos⁡x)sin⁡x(1−cos⁡x)\frac{2(1 - \cos x)}{\sin x(1-\cos x)}
  6. Cancel terms: Cancel out the (1−cos⁡x)(1 - \cos x) terms in the numerator and denominator.2sin⁡x\frac{2}{\sin x}
  7. Recognize final result: Recognize that 2sin⁡x\frac{2}{\sin x} is the same as 2csc⁡x2\csc x.\newline2csc⁡x2\csc x

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