# Inequality involving tangent provided $\tan\theta\geq 1$

If $\tan\theta\geq1$, then $$\sin\theta-\cos\theta\leq\mu(\cos\theta+\sin\theta)\implies\tan\theta\leq\dfrac{1+\mu}{1-\mu}.$$

Why? I get as far as the obvious $$\tan\theta\leq1+\mu(\cos\theta+\sin\theta)$$

• Is $\mu$ any real number? – Alex Jul 10 '13 at 14:38

## 1 Answer

Assume for the moment that we are in quadrant I (i.e., $\sin{\theta} \ge 0$ and $\cos{\theta} \ge 0$). Then

$$\frac{\sin{\theta}-\cos{\theta}}{\sin{\theta}+\cos{\theta}} \le \mu$$

Divide through up and down by $\cos{\theta}$:

$$\frac{\tan{\theta}-1}{\tan{\theta}+1} \le \mu$$

which means that

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

The result follows. You then need to show that this also works in quadrant III.

• @Valentina: you're welcome. I noticed that you asked 6 questions and not accepted any answers. Just so you know, if you get an answer useful to you, you can accept it by clicking on the checkmark to the left. – Ron Gordon Jul 10 '13 at 13:53
• @RonGordon I don't understand your statement that said $\frac{\sin{\theta}-\cos{\theta}}{\sin{\theta}+\cos{\theta}} \le \mu$ . If we assume the equality holds, then we can substitute into the original question and get $\sin \theta - \cos ]theta \le \sin \theta - \cos \theta$ , which is true. But according to your inequality this is the max of the RHS. Shouldn't this be the min, and then your inequality would be $\frac{\sin{\theta}-\cos{\theta}}{\sin{\theta}+\cos{\theta}} \ge \mu$ ? – Ovi Jul 11 '13 at 5:41