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

simplify the expression. \n\n\\( \\sqrt { \\frac { 1 - \\cos 1 3 3 ^ { \\circ } } { 1 + \\cos 1 3 3 ^ { \\circ } } } \\)\n\n\\( \\sqrt { \\frac { 1 - \\cos 1 3 3 ^ { \\circ } } { 1 + \\cos 1 3 3 ^ { \\circ } } } = \\square ^ { \\circ } \\)\n(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 3 ^ { \\circ } } { 1 + \\cos 1 3 3 ^ { \\circ } } } \\)\n\n\\( \\sqrt { \\frac { 1 - \\cos 1 3 3 ^ { \\circ } } { 1 + \\cos 1 3 3 ^ { \\circ } } } = \\square ^ { \\circ } \\)\n(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 half - angle identity

We know the identity (\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}). For (\alpha = 133^{\circ}), the expression (\sqrt{\frac{1-\cos133^{\circ}}{1 + \cos133^{\circ}}}) is equal to (\left|\tan\frac{133^{\circ}}{2}\right|). Since (0^{\circ}<\frac{133^{\circ}}{2}=66.5^{\circ}<90^{\circ}), (\tan66.5^{\circ}>0).

Step2: Calculate the value

(\sqrt{\frac{1-\cos133^{\circ}}{1+\cos133^{\circ}}}=\tan\frac{133^{\circ}}{2}) (\tan66.5^{\circ}\approx2.3)

Answer:

(2.3)