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Soit a une mesure d'un angle orienté. Recopie et exprime chacune des expressions suivantes en fonction de tan a.

Mathematics

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Answers

5 parts
a

−tan⁡(α)-\tan(\alpha)

b

−tan⁡(α)-\tan(\alpha)

c

1tan⁡(α)\frac{1}{\tan(\alpha)}

d

tan⁡(α)\tan(\alpha)

e

−1tan⁡(α)-\frac{1}{\tan(\alpha)}

Step-by-step solution

7 steps

Here are the expressions in terms of tan⁡α\tan \alpha:

Part a· 1 step

  1. Use the property that tan⁡(−x)=−tan⁡(x)\tan(-x) = -\tan(x).

    tan⁡(−α)=−tan⁡(α)\tan(-\alpha) = -\tan(\alpha)

    The expression is −tan⁡(α)\boxed{-\tan(\alpha)}.

Part b

−tan⁡(α)-\tan(\alpha)

  1. Use the property that tan⁡(π−x)=−tan⁡(x)\tan(\pi - x) = -\tan(x).

    tan⁡(π−α)=−tan⁡(α)\tan(\pi - \alpha) = -\tan(\alpha)

    The expression is −tan⁡(α)\boxed{-\tan(\alpha)}.

Part c

1tan⁡(α)\frac{1}{\tan(\alpha)}

  1. Use the property that tan⁡(π2−x)=cot⁡(x)\tan(\frac{\pi}{2} - x) = \cot(x).

  2. Express cot⁡(x)\cot(x) in terms of tan⁡(x)\tan(x).

    tan⁡(π2−α)=cot⁡(α)=1tan⁡(α)\tan\left(\frac{\pi}{2} - \alpha\right) = \cot(\alpha) = \frac{1}{\tan(\alpha)}

    The expression is 1tan⁡(α)\boxed{\frac{1}{\tan(\alpha)}}.

Part d

tan⁡(α)\tan(\alpha)

  1. Use the property that tan⁡(π+x)=tan⁡(x)\tan(\pi + x) = \tan(x).

    tan⁡(π+α)=tan⁡(α)\tan(\pi + \alpha) = \tan(\alpha)

    The expression is tan⁡(α)\boxed{\tan(\alpha)}.

Part e

−1tan⁡(α)-\frac{1}{\tan(\alpha)}

  1. Use the property that tan⁡(π2+x)=−cot⁡(x)\tan(\frac{\pi}{2} + x) = -\cot(x).

  2. Express cot⁡(x)\cot(x) in terms of tan⁡(x)\tan(x).

    tan⁡(π2+α)=−cot⁡(α)=−1tan⁡(α)\tan\left(\frac{\pi}{2} + \alpha\right) = -\cot(\alpha) = -\frac{1}{\tan(\alpha)}

    The expression is −1tan⁡(α)\boxed{-\frac{1}{\tan(\alpha)}}.

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