6.1 \( \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) \( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 6.3 If \( \tan x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} \) and \( m^{2}+\frac{1}{m^{2}}=1 \) Calculate the value of \( x \) without the use of a calculator

In the expression \( \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \), remember the unit circle! The sine and cosine values can be easily retrieved: \( \sin 210^{\circ} = -\frac{1}{2} \), \( \cos 300^{\circ} = \frac{1}{2} \), and \( \tan 240^{\circ} = \sqrt{3} \). With a little simplification, this expression invites you to a numerical dance that leads straight to the answer—just make sure to keep track of those negatives! As for the expression \( [\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \), imagine combining angles like recipes! Using identities, we can sprinkle in \( \sin (-\theta) = -\sin(\theta) \) and \( \cos(360-\theta) = \cos(\theta) \). When we simplify further using \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we're on our way to revealing the ultimate flavor of this trigonometric concoction! Now, let’s tackle \( \tan x = m + \frac{1}{m} \) and \( m^2 + \frac{1}{m^2} = 1 \). From \( m^2 + \frac{1}{m^2} = 1 \), you can cleverly deduce that \( m = 1 \) or \( m = -1 \). This helps identify that \( \tan x = 2 \) or \( \tan x = 0 \). Given the interval \( 90^{\circ} \leq x \leq 270^{\circ} \), the only plausible value comes from \( \tan x = 0 \), leading us gracefully to \( x = 180^{\circ} \). Voilà, that’s the value of \( x \)!