\( 6.1 \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) \( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 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. C

To solve the problem: **Given:** - \( \tan x = m + \frac{1}{m} \) - \( 90^{\circ} \leq x \leq 270^{\circ} \) - \( m^{2} + \frac{1}{m^{2}} = 1 \) **Step-by-Step Solution:** 1. **Express \( \tan x \) in terms of \( m \):** \[ \tan x = m + \frac{1}{m} \] 2. **Use the given equation \( m^{2} + \frac{1}{m^{2}} = 1 \):** \[ m^{2} + \frac{1}{m^{2}} = 1 \] Notice that: \[ \left( m + \frac{1}{m} \right)^2 = m^2 + 2 \cdot m \cdot \frac{1}{m} + \frac{1}{m^2} = m^2 + 2 + \frac{1}{m^2} \] Therefore: \[ \left( m + \frac{1}{m} \right)^2 = 1 + 2 = 3 \implies m + \frac{1}{m} = \pm\sqrt{3} \] So: \[ \tan x = \pm\sqrt{3} \] 3. **Determine \( x \) within the given interval \( 90^{\circ} \leq x \leq 270^{\circ} \):** - **If \( \tan x = \sqrt{3} \):** - The general solution for \( \tan x = \sqrt{3} \) is \( x = 60^{\circ} + k \cdot 180^{\circ} \). - Within \( 90^{\circ} \leq x \leq 270^{\circ} \): \[ x = 240^{\circ} \] - **If \( \tan x = -\sqrt{3} \):** - The general solution for \( \tan x = -\sqrt{3} \) is \( x = 120^{\circ} + k \cdot 180^{\circ} \). - Within \( 90^{\circ} \leq x \leq 270^{\circ} \): \[ x = 120^{\circ} \] 4. **Conclusion:** - The possible values of \( x \) within the specified interval that satisfy the given conditions are: \[ x = 120^{\circ} \quad \text{and} \quad x = 240^{\circ} \] **Final Answer:** The possible values of \( x \) are \(120^\circ\) and \(240^\circ\).