Question

simplify the expression.

\\( \sqrt { \frac { 1 - \cos 1 3 9 ^ { \circ } } { 1 + \cos 1 3 9 ^ { \circ } } } \\)

\\( \sqrt { \frac { 1 - \cos 1 3 9 ^ { \circ } } { 1 + \cos 1 3 9 ^ { \circ } } } = \square ^ { \circ } \\)
(simplify your answer. use integers or decimals for any numbers in the expression. do not include the degree symbol in your answer.)

Explanation:

Step1: Use the half - angle identity

We know the identity \(\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}\). Here \(\alpha = 139^{\circ}\), and \(\frac{\alpha}{2}=\frac{139^{\circ}}{2}=69.5^{\circ}\). Since \(0^{\circ}<69.5^{\circ}<90^{\circ}\), \(\tan\frac{139^{\circ}}{2}>0\).
So \(\sqrt{\frac{1 - \cos139^{\circ}}{1+\cos139^{\circ}}}=\tan\frac{139^{\circ}}{2}\)

Step2: Calculate the value

\(\tan\frac{139^{\circ}}{2}=\tan69.5^{\circ}\approx2.747\)

Answer:

\(2.747\)