For acute $\theta$, write $\cot\theta$ in terms of $\sin\theta$

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For acute $\theta$, write $\cot\theta$ in terms of $\sin\theta$.

I know that's $\cot\theta = \frac{\cos\theta}{\sin\theta}$ but why is the answer $\cot\theta= \frac{\sqrt {1-\sin^2\theta}}{\sin\theta}$?

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1 Answer

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Using the Pythagorean identity:$$\sin^2(\theta)+\cos^2(\theta) = 1$$Subtracting $\sin^2(\theta)$ from both sides:$$\cos^2(\theta) = 1-\sin^2(\theta) $$ Because $\theta$ is acute ($0^\circ < \theta < 90^\circ$) we can take the square root of both sides and keep all of our signs the same ($\sin, \cos, \tan$ are all positive when $\theta$ is acute) so we now have:$$\cos(\theta) = \sqrt{1-\sin^2(\theta)}$$

So from $\cot(\theta) = \dfrac{\cos(\theta)}{\sin(\theta)}$: $$\boxed{\cot(\theta) = \frac{\sqrt{1-\sin^2(\theta)}}{\sin(\theta)}}$$

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