find the exact value of the function.\n\n\\( \\tan \\frac { \\beta } { 2 } \\), given \\( \\tan \\beta =…

Explanation:

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

Since (180^{\circ}<\beta < 270^{\circ}), then (90^{\circ}<\frac{\beta}{2}<135^{\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{1 - \cos\alpha}{\sin\alpha}) or (\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1 + \cos\alpha}}). Another form is (\tan\frac{\beta}{2}=\frac{\sin\beta}{1+\cos\beta}). Given (\tan\beta=\frac{\sqrt{7}}{3}=\frac{\sin\beta}{\cos\beta}), and (\sin^{2}\beta+\cos^{2}\beta = 1). Since (\tan\beta=\frac{\sqrt{7}}{3}) and (180^{\circ}<\beta<270^{\circ}), we know that (\sin\beta=-\frac{\sqrt{7}}{\sqrt{7 + 9}}=-\frac{\sqrt{7}}{4}) and (\cos\beta=-\frac{3}{4}) (because in the third quadrant (\sin\beta<0) and (\cos\beta<0)).

Step3: Substitute into the half - angle formula

(\tan\frac{\beta}{2}=\frac{\sin\beta}{1+\cos\beta}) Substitute (\sin\beta =-\frac{\sqrt{7}}{4}) and (\cos\beta=-\frac{3}{4}) into the formula: [ \begin{align*} \tan\frac{\beta}{2}&=\frac{-\frac{\sqrt{7}}{4}}{1-\frac{3}{4}}\ &=\frac{-\frac{\sqrt{7}}{4}}{\frac{1}{4}}\ &=-\sqrt{7} \end{align*} ]

Answer:

(-\sqrt{7})