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.

Given: \[ \tan x = m + \frac{1}{m} \] and \[ m^2 + \frac{1}{m^2} = 1 \] First, we simplify the equation \( m^2 + \frac{1}{m^2} \): \[ m^2 + \frac{1}{m^2} = 1 \implies \left( m + \frac{1}{m} \right)^2 - 2 = 1 \implies \left( m + \frac{1}{m} \right)^2 = 3 \] This implies: \[ m + \frac{1}{m} = \sqrt{3} \quad \text{or} \quad m + \frac{1}{m} = -\sqrt{3} \] Next, we will consider both cases. **Case 1:** \( m + \frac{1}{m} = \sqrt{3} \) To find \( m \), we can set \( t = m \), leading to: \[ t^2 - \sqrt{3}t + 1 = 0 \] Using the quadratic formula: \[ t = \frac{\sqrt{3} \pm i}{2} \] This means: \[ m = \frac{\sqrt{3}}{2} + i\frac{1}{2} \quad \text{or} \quad m = \frac{\sqrt{3}}{2} - i\frac{1}{2} \] Calculating \( \tan x \): \[ \tan x = \sqrt{3} \] This corresponds to values of \( x \) that satisfy \( \tan x = \sqrt{3} \) within the range \( 90^\circ \leq x \leq 270^\circ\): The angle \( x = 120^\circ \) (in the second quadrant) satisfies this. **Case 2:** \( m + \frac{1}{m} = -\sqrt{3} \) Similarly, we set: \[ t^2 + \sqrt{3}t + 1 = 0 \] Using the quadratic formula again: \[ t = \frac{-\sqrt{3} \pm i}{2} \] This implies: \[ m = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \quad \text{or} \quad m = -\frac{\sqrt{3}}{2} - i\frac{1}{2} \] Calculating \( \tan x \): \[ \tan x = -\sqrt{3} \] This corresponds to values of \( x \) that satisfy \( \tan x = -\sqrt{3} \) within the range \( 90^\circ \leq x \leq 270^\circ\): The angle \( x = 240^\circ \) (in the third quadrant) satisfies this. Finally, the possible values of \( x \) are: \[ x = 120^\circ \quad \text{and} \quad x = 240^\circ \] Thus, both values are valid depending on the \( m \) chosen. If you want just one, you can choose either: \[ \boxed{120^\circ} \quad \text{or} \quad \boxed{240^\circ} \]