How would you verify 2cosø/2sin ø + sinø/(1 +cos) ø= csc ø?

Question

Please help. I have a test tomorrow.

Answer 1

Work on the Left hand side

2cosø/2sin ø + sinø/(1 +cosø)

Find a common denominator - ((2sinø(1 +cosø) )

(2cosø(1 +cosø))/((2sinø(1 +cosø))+(2sinø*sinø)/((1 +cosø) 2sinø)


(2cosø + 2cos^2ø)/ )/((2sinø(1 +cosø) ) + (2sinø*sinø)/ )/((2sinø(1 +cosø) )

(2cosø + 2cos^2ø + 2 sin^2ø) / ((2sinø(1 +cosø) )

2cosø + 2(cos^2ø + sin^2ø) / ((2sinø(1 +cosø) )

2cosø + 2 / ((2sinø(1 +cosø) )

2(cos ø + 1) / ((2sinø(1 +cosø) )

2 / 2 sin ø

1 / sin ø

csc ø

Answer 2

2cosø/2sin ø + sinø/(1 +cosø)= cosec ø ? this is the correct form
Proof :
= cosø/sin ø + sinø/(1 +cosø) // 2/2=1
= [cosø(1 +cosø)+sin ø*sin ø] / sinø(1 +cosø) // cross multiplying
= [cosø+(cosø)^2+(sin ø)^2] / sinø(1+cosø) // opening bracket
= [1+cosø] / sinø(1 +cosø) // (sin ø)^2+(cosø)^2=1
= 1 / sinø
= cosecø

Answer 3

2cosø/2sin ø + sinø/(1 +cosø) = csc ø
cosø(1+cosø)/[sin ø(1 +cosø)] + sin^2ø/[sinø(1 +cosø)] = (1 +cosø)/[sinø(1 +cosø)]
So cosø +cos^ø +sin^ø =1+cosø
cosø +1 =1+cosø
cosø = cosø

Answer 4

2cost/2sint + sint/(1 + cost) = csct

Choose the left hand side.

LHS = 2cost/2sint + sint / (1 + cost)

Note that the 2s will cancel themselves out, leaving

LHS = cos(t)/sin(t) + sin(t)/(1 + cos(t))

Put them under a common denominator.

LHS = cos(t)[1 + cos(t)]/[(sin(t)) (1 + cos(t))] +
sin(t)sin(t)/[sin(t) (1 + cos(t))]

Merge them as one unit.

LHS = {cos(t)[1 + cos(t)] + sin(t)sin(t)}/[sin(t) (1 + cos(t))]

Now, simplify the numerator

LHS = {cos(t) + cos^2(t) + sin^2(t)} / [sin(t) (1 + cos(t))]

Change cos^2(t) + sin^2(t) to 1.

LHS = {1 + cos(t)} / [sin(t) (1 + cos(t))]

We have cancelling terms; (1 + cos(t)) on the top and bottom.

LHS = 1/sin(t)
LHS = csc(t) = RHS