1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Focus on triangles DFC and DFE:
- They are both right triangles because DF is perpendicular to EC
- DC=DE
- DF is a common side
So, the two triangles are congruent. It follows that CF=FE, and in particular
And since x=3, the length of EF is
Answer: 67.353
Step-by-step explanation:
60.000
+07.000
+00.300 Writing the numbers like this helps me keep track of them!
+00.050 It lets me see what place each number is in relative to the others
+00.003
67.353
