Choose two of the angles on ∆ABC, and locate the line segment between them. Draw a new line segment,DE , parallel to the line se
gment you located on ∆ABC. You can draw of any length and place it anywhere on the coordinate plane, but not on top of ∆ABC. From points D and E, create an angle of the same size as the angles you chose on ∆ABC. Then draw a ray from D and a ray from E through the angles such that the rays intersect. You should now have two angles that are congruent to the angles you chose on ∆ABC. Label the point of intersection of the two rays F, and draw ∆DEF by creating a polygon through points D, E, and F.
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Answer:
See detail below.
Step-by-step explanation:
A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:
(cosxtanx-tanx+2cosx-2)/(tanx+2)
Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.
Now to the actual problem: \we can make the following transformations on the left hand side:
which is shown to be the same as the right hand side, which was to be shown.