Find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$.

Question

Given that $\tan$ $\alpha = p$, where $\alpha$ is acute, find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$.

Answer 1

Hint: use the identity $$\tan(\theta+\phi)\equiv \frac{\tan(\theta)+\tan(\phi)}{1-\tan(\theta)\tan(\phi)}.$$

Substitute the appropriate values of $\theta$ and $\phi$ into the above identity.

For the first part, note that $\tan(-\theta)\equiv -\tan(\theta)$ (since $\tan$ is an odd function).

Answer 2

• $\tan(-\alpha)=-\tan(\alpha)$ (Prove it!). Thus, $\tan(-\alpha)=-p$.

• $\tan(\pi-\alpha)=-\tan(\alpha)$ (Prove it!). Thus, $\tan(\pi-\alpha)=-p$.

• $\tan(\frac{\pi}{2}-\alpha)=\cot(\alpha)$ (Prove it!). Thus, $\tan(\frac{\pi}{2}-\alpha)=p^{-1}$.

Answer 3

Hint: Use the following facts:

• $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$
• $\sin(-\alpha) = -\sin(\alpha)$
• $\cos(-\alpha) = \cos(\alpha)$
• $\sin(\pi-\alpha) = \sin(\alpha)$
• $\cos(\pi-\alpha) = -\cos(\alpha)$
• $\sin(\frac{\pi}{2}-\alpha) = \cos(\alpha)$
• $\cos(\frac{\pi}{2}-\alpha) = \sin(\alpha)$.

Answer 4

$$\cos\dfrac{\pi}{2+\alpha} = - \sin \alpha \\ \sin\dfrac{\pi}{2+\alpha} = \cos \alpha$$