simplify the expression. \n\n sqrt { \frac { 1 - cos 133 ^ { circ } } { 1 + cos 133 ^ { circ } } } \n\n sqrt…

simplify the expression. \n\n sqrt { \frac { 1 - cos 133 ^ { circ } } { 1 + cos 133 ^ { circ } } } \n\n sqrt { \frac { 1 - cos 133 ^ { circ } } { 1 + cos 133 ^ { 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 133 ^ { circ } } { 1 + cos 133 ^ { circ } } } \n\n sqrt { \frac { 1 - cos 133 ^ { circ } } { 1 + cos 133 ^ { 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 that (\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}). Here (\alpha = 133^{\circ}), and since (0^{\circ}<\frac{133^{\circ}}{2}=66.5^{\circ}<90^{\circ}), (\tan\frac{133^{\circ}}{2}>0). So (\sqrt{\frac{1 - \cos133^{\circ}}{1+\cos133^{\circ}}=\tan\frac{133^{\circ}}{2}).

Step2: Calculate the value

(\frac{133^{\circ}}{2}=66.5)

Answer:

(66.5)