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 first analyze the given equation \( m^2 + \frac{1}{m^2} = 1 \). We can rewrite this as: \[ m^2 + \frac{1}{m^2} = \left( m + \frac{1}{m} \right)^2 - 2 \] Let \( y = m + \frac{1}{m} \). Thus, we have: \[ y^2 - 2 = 1 \implies y^2 = 3 \implies y = \sqrt{3} \text{ or } y = -\sqrt{3} \] The values of \( m + \frac{1}{m} \) can be written as \( \tan x \), thus: \[ \tan x = \sqrt{3} \quad \text{or} \quad \tan x = -\sqrt{3} \] Now, we need to find the angles \( x \) within the interval \( 90^\circ \leq x \leq 270^\circ \). 1. For \( \tan x = \sqrt{3} \): - This occurs at \( x = 60^\circ \) in the first quadrant but since \( 60^\circ \) is not within our specified range, we look for the corresponding angle in the third quadrant, which is: \[ x = 180^\circ + 60^\circ = 240^\circ. \] 2. For \( \tan x = -\sqrt{3} \): - This occurs at \( x = 120^\circ \) (second quadrant) and \( x = 300^\circ \) (fourth quadrant). Since \( 120^\circ \) is within the range, we also check for \( 240^\circ \) already found. Thus, the possible angles based on the ranges are \( 120^\circ \) and \( 240^\circ \). But since \( \tan x \) must equal \( m + \frac{1}{m} \): Since all possibilities of \( m + \frac{1}{m} \) have been assigned to \( \tan x \) and checked for the valid quadrants, we confirm: \(\tan x = \sqrt{3} \implies x = 240^\circ\) because in the specific given range, only \( 240^\circ \) fits all conditions. Therefore, the solution is: \[ \boxed{240^\circ} \]