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Please help simplify expression

\begin{align}\cot\alpha\ (1-\cos2\alpha)\end{align} I tried to solve through this way: First dividing and multiplying both parts by $2$

\begin{align}\cot\alpha\ (1-\cos2\alpha)= 2\cot\alpha\sin^2\alpha\end{align}

Second reducing the degree and sin of cot \begin{align}2\cos\alpha\sin\alpha=\sin 2\alpha\end{align} But in book there is $-4\sin\alpha$

Where is my mistake?

Harish Chandra Rajpoot
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Krutya
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    Your answer is correct, the book's wrong. – Nikunj Jul 04 '20 at 17:24
  • You can tell the book is wrong because $1-\cos 2\alpha$ is never negative and, for (say) $\alpha$ in the first quadrant, $\cot \alpha$ is positive. Therefore, the initial expression is non-negative for first-quadrant angles, so that any equivalent expression should be, too; however, $-4\sin\alpha$ is negative for such angles. – Blue Jul 04 '20 at 21:33

1 Answers1

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Your answer is correct. The answer $-4\sin\alpha$ in your book isn't correct. Substitute $\cot\alpha=\frac{\cos\alpha}{\sin\alpha}$ to get to correct answer $$2\cot\alpha\sin^2\alpha=2\frac{\cos\alpha}{\sin\alpha}\sin^2\alpha=2\sin\alpha\cos\alpha=\color{blue}{\sin2\alpha}$$

Harish Chandra Rajpoot
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