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(1-cos alpha)/(sin alpha)=2csc alpha

1−cos⁡αsin⁡α=2csc⁡α \frac{1-\cos \alpha}{\sin \alpha}=2 \csc \alpha

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Full solution

Q. 1−cos⁡αsin⁡α=2csc⁡α \frac{1-\cos \alpha}{\sin \alpha}=2 \csc \alpha
  1. Express in terms of sin: Express csc⁡(α)\csc(\alpha) in terms of sin⁡(α)\sin(\alpha) as csc⁡(α)=1sin⁡(α)\csc(\alpha) = \frac{1}{\sin(\alpha)}.
  2. Substitute with 1/sin⁡:1/\sin: Substitute csc⁡(α)\csc(\alpha) with 1/sin⁡(α)1/\sin(\alpha) in the right side of the equation to get 2/sin⁡(α)2/\sin(\alpha).
  3. Multiply to eliminate denominator: Multiply both sides of the equation by sin⁡(α)\sin(\alpha) to eliminate the denominator on the left side. This gives us sin⁡(α)×(1−cos⁡(α))/sin⁡(α)=2sin⁡(α)×sin⁡(α)\sin(\alpha) \times (1-\cos(\alpha))/\sin(\alpha) = \frac{2}{\sin(\alpha)} \times \sin(\alpha).
  4. Simplify by canceling terms: Simplify the left side by canceling out sin⁡(α)\sin(\alpha) in the numerator and the denominator, which leaves us with 1−cos⁡(α)1-\cos(\alpha). On the right side, sin⁡(α)\sin(\alpha) cancels out as well, leaving us with 22.
  5. Identify math error: We now have the simplified equation 1−cos⁡(α)=21-\cos(\alpha) = 2. However, this is not correct as the original equation was (1−cos⁡(α))/sin⁡(α)=2csc⁡(α)(1-\cos(\alpha))/\sin(\alpha) = 2\csc(\alpha), and we cannot simply cancel out sin⁡(α)\sin(\alpha) on the left side without affecting the right side of the equation. This is a math error.

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