\( 6 .(9-2.6) \) Biaqucante: \( \left(-30^{\circ}\right): \cos \left(-60^{\circ}\right)+\operatorname{tg}\left(-30^{\circ}\right) \cdot \operatorname{ctg}^{2}\left(-60^{\circ}\right) \) \( \begin{array}{ll}\text { A) }-1-\frac{\sqrt{9}}{9} & \text { B) } 1-\frac{\sqrt{3}}{9} \\ \text { () }-1+\frac{\sqrt{3}}{9} & \text { D) } 1+\frac{\sqrt{3}}{9}\end{array} \)

To tackle the expression \( 6 .(9-2.6) \), first we simplify it. We calculate \( 9 - 2.6 = 6.4 \), then multiply that by 6 to get \( 6 \times 6.4 = 38.4 \). Now, let's break down the trigonometric values involved in the second part. We know that \( \cos(-60^{\circ}) = \frac{1}{2} \) and \( \tan(-30^{\circ}) = -\frac{1}{\sqrt{3}} \). The cosecant squared for \( -60^{\circ} \) is \( \cot^{2}(-60^{\circ}) = 3 \). Plugging everything into the expression gives us \( (-30^{\circ}) : \cos(-60^{\circ}) + \tan(-30^{\circ}) \cdot \cot^{2}(-60^{\circ}) = -1 + \left(-\frac{1}{\sqrt{3}}\right)3 = -1 - \sqrt{3} \). Now we have: \( \begin{array}{ll}\text { A) } -1-\frac{\sqrt{9}}{9} & \text { B) } 1-\frac{\sqrt{3}}{9} \\ \text { C) } -1+\frac{\sqrt{3}}{9} & \text { D) } 1+\frac{\sqrt{3}}{9}\end{array} \) After evaluating the possible answers, option A has the most fitting result based on the calculations! In any case, understanding trigonometric identities can truly help in solving complex problems easily. It’s all about knowing your angles and their respective values in the unit circle! Fun fact: Did you know that trigonometric functions are not just for math class? They're used in everything from computer graphics to music theory, helping create and visualize sound waves and animations. An essential toolkit for many fields!