simplify the expression. \n\n\\( \\sqrt { \\frac { 1 - \\cos 1 3 9 ^ { \\circ } } { 1 + \\cos 1 3 9 ^ {…

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

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

Answer

Explanation:

Step1: Use the trigonometric 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}=\sqrt{\frac{1 - \cos139^{\circ}}{1+\cos139^{\circ}}})

Answer:

(69.5)