Simplify the following expressions without using a calculator, \( 6.1 \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) \( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 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

To simplify the expression \( 6.1 \): First, we find the values of the trigonometric functions: - \( \sin 210^{\circ} = -\frac{1}{2} \) - \( \cos 300^{\circ} = \frac{1}{2} \) - \( \tan 240^{\circ} = \frac{1}{\sqrt{3}} \) - \( \cos 120^{\circ} = -\frac{1}{2} \) - \( \tan 150^{\circ} = -\frac{1}{\sqrt{3}} \) - \( \sin 330^{\circ} = -\frac{1}{2} \) Now substituting these values into the expression: \[ 6.1 = 6.1 \cdot \frac{-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{3}}}{-\frac{1}{2} \cdot -\frac{1}{\sqrt{3}} \cdot -\frac{1}{2}} \] The numerator simplifies to \( -\frac{1}{8\sqrt{3}} \) and the denominator simplifies to \( -\frac{1}{8\sqrt{3}} \): \[ 6.1 = 6.1 \cdot 1 = 6.1 \] Now to simplify \( 6.2 \): Using the identities: - \( \sin(-\theta) = -\sin(\theta) \) - \( \cos(360 - \theta) = \cos(\theta) \) - \( \cos(90 - \theta) = \sin(\theta) \) and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) We replace these in \( 6.2 \): \[ 6.2 = [-\sin\theta + \cos\theta][\sin\theta + \frac{\sin\theta \cdot \cos\theta}{\sin\theta}] \] \[ = [-\sin\theta + \cos\theta][\sin\theta + \cos\theta] \] Applying the formula for the product of binomials: \[ = (-\sin^2\theta + \cos^2\theta) + (-\sin\theta \cos\theta + \sin\theta \cos\theta) = \cos^2\theta - \sin^2\theta \] So, we can conclude the simplified expressions as \( 6.1 = 6.1 \) and \( 6.2 = \cos^2\theta - \sin^2\theta \). For \( 6.3 \): Given \( m^2 + \frac{1}{m^2} = 1 \), we let \( t = m + \frac{1}{m} \): \[ t^2 = m^2 + 2 + \frac{1}{m^2} = 1 + 2 \implies t^2 = 3 \implies t = \sqrt{3} \text{ or } -\sqrt{3} \] Now, since \( \tan x = m + \frac{1}{m} \), 1. \( \tan x = \sqrt{3} \) corresponds to \( x = 60^{\circ} + k \cdot 180^{\circ} \) 2. \( \tan x = -\sqrt{3} \) corresponds to \( x = 120^{\circ} + k \cdot 180^{\circ} \) In the interval \( 90^{\circ} \leq x \leq 270^{\circ} \): - The solutions for \( x = \sqrt{3} \) fall into the quadrant of \( 60^{\circ} + 180^{\circ} \), giving us \( 240^{\circ} \). - The solution for \( -\sqrt{3} \) is \( 120^{\circ} \). Finally, the valid \( x \) is \( 120^{\circ} \) and \( 240^{\circ} \), but since it is specified \( 90^{\circ} \leq x \leq 270^{\circ} \), we confirm that \( x = 120^{\circ} \) and \( x = 240^{\circ} \).