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
Answer:
A. (55 x 5) - (40 x 5)
Step-by-step explanation:
You are solving how much miles (further along) would the second car be after 5 hours.
The first car averages 40 miles per hour. 5 hours later, it will have averaged about 200 miles in 5 hours (40 x 5 = 200).
The second car averages 55 miles per hour. 5 hours later, it will have averaged about 275 miles in 5 hours (55 x 5 = 275)
Subtract: 275 - 200 = 75
The second car would have averaged 75 more miles than the first car.
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Answer:
slope = 5/8
Step-by-step explanation:
use the slope formula to find the slope: 
