Question 2 1 pts Why is v5 + 5 tan2 0 = v5 sec- 0? Because 1 + tan- 0 = sec2 0. O Because vtan d = sec 0 O Because sec- 0 = cos 0 Question 3 1 pts Why is v5 sec2 0 = 15 sec 0? O Because Va' . by = avb Hint: don't pick this one!) O Because Va- + bl = c Because VA . B = VA . VB Question 4 1 pts Why is VS sec ! = sec 0? V5 sec 8 O Because sec # = = O Because sec- 0 = (sec 0) (sec 0) Because sec- 0 = tan- 0mon 6 ".I'li'hwr does the substitution statement I = 1.3 ten 5' correspond to the middle reference triangle shown here? In our substitution statement, =J3. In a right triangle,tan H = fad]. Our substitution statement can be written as ten 3 = %. Therefore, the side adjacent to angle 9 Is labeled by :3 {representing a = 1,33]. In the appropriate reference triangle. And, the side angle 3 Is labeled by , In the appropriate reference triangle. Question 7 In a right triangle, cos 0 = kyp Also, sec 0 = COS B Therefore, in a right triangle, sec 0 = - hyp adj Using the reference triangle shown above, and a = v5, which expression is equivalent to sec 0? O sec = seco = Vets 1/5 O sec e = Question 8 Which of the following is an acceptable answer, when we are asked to evaluate / =dx? V5+x + +c O sec Vx-+5 +C O In| sec # + tan #| + C