\( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) 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 \( 6.2[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \): First, remember that \( \sin(-\theta) = -\sin(\theta) \) and \( \cos(360^\circ - \theta) = \cos(\theta) \). So, you have: \[ \sin(-\theta) + \cos(360^\circ - \theta) = -\sin(\theta) + \cos(\theta). \] Next, know that \( \cos(90^\circ - \theta) = \sin(\theta) \). Thus: \[ \cos(90^\circ - \theta) + \frac{\sin \theta}{\tan \theta} = \sin(\theta) + \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \sin(\theta) + \cos(\theta). \] Combining these two results gives: \[ 6.2\left[-\sin(\theta) + \cos(\theta)\right]\left[\sin(\theta) + \cos(\theta)\right]. \] Now for the second part with \( \tan x = m + \frac{1}{m} \) and \( m^2 + \frac{1}{m^2} = 1 \). Using the identity \( m^2 + \frac{1}{m^2} = (m + \frac{1}{m})^2 - 2 \), we find \( m + \frac{1}{m} = 2 \) or \( -2 \). From this, if \( \tan x = 2 \) or \( -2 \) and considering the ranges provided (\( 90^\circ \leq x \leq 270^\circ \)), we recognize \( \tan x \) is negative in the second and third quadrants. The only possible value for \( \tan x = -2 \), occurs at \( x = 180^\circ + \tan^{-1}(2) \) which is approximately \( 180^\circ + 63.43^\circ \approx 243.43^\circ \). Thus, the calculated value of \( x \) is: \[ x \approx 243.43^\circ. \]