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Vesnalui [34]
3 years ago
14

Cosx+1/sin^3x=cscx/1-cosx

Mathematics
1 answer:
ANTONII [103]3 years ago
8 0
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).

 </span>
<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).

 </span>
<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.

 </span>
<span>I hope this helps! </span>
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Evaluate the following expressions if ×=2,y=3 and Z=4. a.4×+×-3over ×+10 b.2y-5yover y+5. c.yz-×. d.9+y+zover y+2​
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2) Fill in an angle measure greater than 100 degrees for the missing angle in yellow below. You will fill
timama [110]

Answer:

m∠a = 110°

m∠b = 110°

m∠c = 70°

m∠d = 70°

m∠i = 110°

m∠h = 110°

m∠j = 70°

m∠g = 110°

m∠f = 70°

m∠e = 70°

m∠k = 57°

m∠m = 70°

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Step-by-step explanation:

Let the measure of the yellow angle = 110°

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m∠b = m∠a = 110° by opposite interior angles of a parallelogram

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m∠d = m∠c = 70° by opposite interior angles of a parallelogram

m∠i and m∠c are supplementary angles, therefore, m∠i = 110°

m∠h = m∠i = 110° by vertically opposite angles

m∠j = m∠c = 70° by vertically opposite angles

m∠g and m∠d are supplementary angles, therefore, m∠g = 110°

m∠f = m∠d = 70° by alternate interior angles

m∠f = m∠e = 70° by vertically opposite angles

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m∠b and m∠m are supplementary angles, therefore, m∠m = 70°

m∠l and 13° are supplementary angles, therefore, m∠l = 167°

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3 years ago
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