<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).
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<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).
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<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.
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<span>I hope this helps! </span>
Answer:
if this is a fraction, then negative 5 over a⁶
All you have to do is divide the left side of the ratio by 5 and the right side of the ratio by 7. The one that comes out even on both sides is the correct one.
Answer:
4k = 24, k = 6
Step-by-step explanation:
4k = 24
4k / 4 = 24 / 4
k = 6
brainliest pls xd
Answer:A) 24 waysB) 4 waysStep-by-step explanation:a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways
Step-by-step explanation:
a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways(hope this helps:)