\( 6.1 \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) \( 62[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 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 calcula

Let's solve the problem step by step. **Given:** 1. \( \tan x = m + \frac{1}{m} \) 2. \( m^2 + \frac{1}{m^2} = 1 \) 3. \( 90^{\circ} \leq x \leq 270^{\circ} \) **Objective:** Find the value of \( x \) without using a calculator. --- ### Step 1: Simplify \( m^2 + \frac{1}{m^2} = 1 \) Notice that: \[ \left( m + \frac{1}{m} \right)^2 = m^2 + 2 \cdot m \cdot \frac{1}{m} + \frac{1}{m^2} = m^2 + 2 + \frac{1}{m^2} \] Given \( m^2 + \frac{1}{m^2} = 1 \): \[ \left( m + \frac{1}{m} \right)^2 = 1 + 2 = 3 \] Taking the square root: \[ m + \frac{1}{m} = \pm \sqrt{3} \] ### Step 2: Substitute Back into \( \tan x \) We have: \[ \tan x = m + \frac{1}{m} = \pm \sqrt{3} \] ### Step 3: Determine Possible Values of \( x \) in the Given Range We need to find \( x \) such that \( 90^{\circ} \leq x \leq 270^{\circ} \) and \( \tan x = \pm \sqrt{3} \). 1. **When \( \tan x = \sqrt{3} \):** - The general solution for \( \tan x = \sqrt{3} \) is \( x = 60^{\circ} + k \cdot 180^{\circ} \), where \( k \) is an integer. - Within \( 90^{\circ} \leq x \leq 270^{\circ} \), the valid solution is: \[ x = 240^{\circ} \] 2. **When \( \tan x = -\sqrt{3} \):** - The general solution for \( \tan x = -\sqrt{3} \) is \( x = 120^{\circ} + k \cdot 180^{\circ} \), where \( k \) is an integer. - Within \( 90^{\circ} \leq x \leq 270^{\circ} \), the valid solution is: \[ x = 120^{\circ} \] ### Conclusion The values of \( x \) that satisfy the given conditions are: \[ \boxed{120^{\circ} \text{ and } 240^{\circ}} \]