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 solve the given problem, we start with the equation \( m^2 + \frac{1}{m^2} = 1 \). We can manipulate this equation as follows: 1. Let \( m^2 = y \); thus, the equation becomes: \[ y + \frac{1}{y} = 1. \] Multiply both sides by \( y \) to eliminate the fraction: \[ y^2 - y + 1 = 0. \] 2. To find the solutions for \( y \), we use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{-3}}{2}. \] This simplifies to: \[ y = \frac{1 \pm i\sqrt{3}}{2}. \] Thus, \( m^2 = \frac{1 \pm i\sqrt{3}}{2} \). 3. We find \( m = \sqrt{\frac{1 + i\sqrt{3}}{2}} \) and \( m = \sqrt{\frac{1 - i\sqrt{3}}{2}} \). To compute \( \tan x \), we substitute into the equation \( \tan x = m + \frac{1}{m} \). 4. Let's compute \( m + \frac{1}{m} \): - First, we need \( \frac{1}{m} = \sqrt{\frac{2}{1 + i\sqrt{3}}} \) and we can use conjugation to simplify: \[ \frac{1}{m} = \frac{1}{\sqrt{\frac{1 + i\sqrt{3}}{2}}} = \sqrt{\frac{2}{1 + i\sqrt{3}}} \cdot \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \sqrt{\frac{2(1 - i\sqrt{3})}{1 + 3}} = \sqrt{\frac{2(1 - i\sqrt{3})}{4}} = \frac{1 - i\sqrt{3}}{2}. \] 5. Adding \( m \) plus \( \frac{1}{m} \): \[ m + \frac{1}{m} = \sqrt{\frac{1 + i\sqrt{3}}{2}} + \sqrt{\frac{1 - i\sqrt{3}}{2}}. \] This value can be shown to be equal to \( i \). 6. Recall from trigonometric identities, if \( \tan x = i \), then: \[ x = \tan^{-1}(i). \] Mathematically, this occurs at \( x = 90^\circ + k \cdot 180^\circ \) for \( k \in \mathbb{Z} \). 7. Given the domain \( 90^\circ \leq x \leq 270^\circ \): - \( k=0 \) gives \( x=90^\circ \) - \( k=1 \) gives \( x=270^\circ \) Thus, since we need the angle where \( \tan x \) gives a specific imaginary number, the only valid range will provide \( 90^\circ \). Therefore, the solution to the problem is: \[ \boxed{90^\circ}. \]