find the exact value of the function.\n\n$\\tan \\frac{\\beta}{2}$, given $\\tan \\beta=-\\frac{\\sqrt{11}}{5…

Explanation:

Step1: Determine the quadrant of (\frac{\beta}{2})

Since (270^{\circ}<\beta < 360^{\circ}), then (135^{\circ}<\frac{\beta}{2}<180^{\circ}). So (\frac{\beta}{2}) is in the second quadrant, and (\tan\frac{\beta}{2}<0).

Step2: Use the half - angle formula for tangent

The half - angle formula for tangent is (\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1 + \cos\alpha}) or (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}). Also, we know that (\tan\beta=-\frac{\sqrt{11}}{5}), and (\tan\beta=\frac{\sin\beta}{\cos\beta}), and (\sin^{2}\beta+\cos^{2}\beta = 1). Since (\tan\beta=-\frac{\sqrt{11}}{5}=\frac{\sin\beta}{\cos\beta}), we have (\sin\beta=-\frac{\sqrt{11}}{5}\cos\beta). Substitute into (\sin^{2}\beta+\cos^{2}\beta = 1): (\left(-\frac{\sqrt{11}}{5}\cos\beta\right)^{2}+\cos^{2}\beta=1) (\frac{11}{25}\cos^{2}\beta+\cos^{2}\beta=1), (\frac{11 + 25}{25}\cos^{2}\beta=1), (\cos^{2}\beta=\frac{25}{36}). Since (270^{\circ}<\beta < 360^{\circ}), (\cos\beta=\frac{5}{6}) (because in the fourth quadrant (\cos) is positive), (\sin\beta=-\frac{\sqrt{11}}{6}) (using (\sin^{2}\beta+\cos^{2}\beta = 1) and (\sin\beta<0) for (270^{\circ}<\beta < 360^{\circ})) Using the half - angle formula (\tan\frac{\beta}{2}=\frac{1-\cos\beta}{\sin\beta}) Substitute (\cos\beta=\frac{5}{6}) and (\sin\beta=-\frac{\sqrt{11}}{6}) (\tan\frac{\beta}{2}=\frac{1-\frac{5}{6}}{-\frac{\sqrt{11}}{6}}=\frac{\frac{1}{6}}{-\frac{\sqrt{11}}{6}}=-\frac{1}{\sqrt{11}}=-\frac{\sqrt{11}}{11})

Answer:

(-\frac{\sqrt{11}}{11})