If $\sin\theta-\cos\theta\leq\mu (\cos\theta + \sin\theta)$, then $\tan\theta \leq \frac{1+\mu}{1-\mu}$

Question

In a text I'm reading, an implication is made:

$$\sin\theta-\cos\theta\leq\mu (\cos\theta + \sin\theta) \quad\implies\quad \tan\theta \leq \frac{1+\mu}{1-\mu}$$

I tried using some trigonometric identities to reproduce this result, but it seems I'm not familiar enough with them.

How was this implication made?

Answer 1

If you divide both sides of the first inequality by $\cos \theta$, you obtain:

$$\frac {\sin \theta - \cos \theta} {\cos \theta} = \tan \theta - 1$$

$$\frac {\mu (\cos \theta + \sin \theta)} {\cos \theta} = \mu ( 1 + \tan \theta)$$

So, assuming $\cos \theta > 0$, from the first inequality you obtain:

$$\tan \theta - 1 \le \mu (1 + \tan \theta) \quad\implies\quad (1 - \mu) \tan \theta \le 1 + \mu$$

Then, assuming $1 - \mu > 0$, you obtain: $$\tan \theta \le \frac {1 + \mu} {1 - \mu}$$

If $\cos \theta < 0$, then by dividing both sides of the first inequality by $\cos \theta$, you must flip the sign. Therefore, the final inequality will be true provided that $1 - \mu < 0$ holds instead.

Answer 2

$$\mu\ge\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$$

$$1+\mu\ge\dfrac{2\sin\theta}{\sin\theta+\cos\theta}$$

$$1-\mu\le\dfrac{2\cos\theta}{\sin\theta+\cos\theta}$$

$$\dfrac{1+\mu}{1-\mu}\ge \tan\theta$$