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

Let's solve each problem step by step. --- ## **6.1** \[ \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \] **Step-by-Step Solution:** 1. **Evaluate each trigonometric function:** - \(\sin 210^{\circ}\): \(210^{\circ} = 180^{\circ} + 30^{\circ}\), so \(\sin 210^{\circ} = -\sin 30^{\circ} = -\frac{1}{2}\). - \(\cos 300^{\circ}\): \(300^{\circ} = 360^{\circ} - 60^{\circ}\), so \(\cos 300^{\circ} = \cos 60^{\circ} = \frac{1}{2}\). - \(\tan 240^{\circ}\): \(240^{\circ} = 180^{\circ} + 60^{\circ}\), so \(\tan 240^{\circ} = \tan 60^{\circ} = \sqrt{3}\). - \(\cos 120^{\circ}\): \(120^{\circ} = 180^{\circ} - 60^{\circ}\), so \(\cos 120^{\circ} = -\cos 60^{\circ} = -\frac{1}{2}\). - \(\tan 150^{\circ}\): \(150^{\circ} = 180^{\circ} - 30^{\circ}\), so \(\tan 150^{\circ} = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}\). - \(\sin 330^{\circ}\): \(330^{\circ} = 360^{\circ} - 30^{\circ}\), so \(\sin 330^{\circ} = -\sin 30^{\circ} = -\frac{1}{2}\). 2. **Substitute the values into the expression:** \[ \frac{(-\frac{1}{2}) \cdot (\frac{1}{2}) \cdot \sqrt{3}}{(-\frac{1}{2}) \cdot (-\frac{1}{\sqrt{3}}) \cdot (-\frac{1}{2})} \] 3. **Simplify the numerator and the denominator:** - **Numerator:** \[ -\frac{1}{2} \times \frac{1}{2} \times \sqrt{3} = -\frac{\sqrt{3}}{4} \] - **Denominator:** \[ -\frac{1}{2} \times -\frac{1}{\sqrt{3}} \times -\frac{1}{2} = -\frac{1}{4\sqrt{3}} \] 4. **Compute the overall fraction:** \[ \frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}} = \frac{\sqrt{3}}{1} \times \sqrt{3} = 3 \] **Answer:** \[ \boxed{3} \] --- ## **6.2** \[ [\sin (-\theta) + \cos (360^{\circ} - \theta)] \left[ \cos (90^{\circ} - \theta) + \frac{\sin \theta}{\tan \theta} \right] \] **Step-by-Step Solution:** 1. **Simplify each trigonometric function:** - \(\sin(-\theta) = -\sin \theta\) - \(\cos(360^{\circ} - \theta) = \cos \theta\) - \(\cos(90^{\circ} - \theta) = \sin \theta\) - \(\frac{\sin \theta}{\tan \theta}\): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \implies \frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta \] 2. **Substitute the simplified forms back into the expression:** \[ [-\sin \theta + \cos \theta] \left[ \sin \theta + \cos \theta \right] \] 3. **Recognize the pattern:** This is of the form \((a + b)(a - b) = a^2 - b^2\). 4. **Apply the identity:** \[ (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - \sin^2 \theta \] 5. **Use the double-angle identity:** \[ \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \] **Answer:** \[ \boxed{\cos 2\theta} \] --- ## **Calculating \(x\) Given:** \[ \tan x = m + \frac{1}{m}, \quad 90^{\circ} \leq x \leq 270^{\circ} \] \[ m^2 + \frac{1}{m^2} = 1 \] **Step-by-Step Solution:** 1. **Given:** \[ m^2 + \frac{1}{m^2} = 1 \] 2. **Let’s find \(\tan x\):** \[ \tan x = m + \frac{1}{m} \] Square both sides to relate with the given equation: \[ (\tan x)^2 = \left(m + \frac{1}{m}\right)^2 = m^2 + 2 + \frac{1}{m^2} \] Substitute the given \(m^2 + \frac{1}{m^2} = 1\): \[ (\tan x)^2 = 1 + 2 = 3 \implies \tan x = \pm\sqrt{3} \] 3. **Determine the possible values for \(x\) in the given interval:** - The interval \(90^{\circ} \leq x \leq 270^{\circ}\) spans the **2nd** and **3rd** quadrants. - **In the 2nd Quadrant (90° to 180°):** \(\tan x\) is **negative**. Thus, \(\tan x = -\sqrt{3}\). \[ x = 180^{\circ} - 60^{\circ} = 120^{\circ} \] - **In the 3rd Quadrant (180° to 270°):** \(\tan x\) is **positive**. Thus, \(\tan x = \sqrt{3}\). \[ x = 180^{\circ} + 60^{\circ} = 240^{\circ} \] **Answer:** \[ \boxed{120^{\circ} \text{ or } 240^{\circ}} \]