Trig final problem.?

Question

express (3scs(x))/(5csc(x)-6(cot(X)squared))

its (3CscX)/(5CscX-6(CotX)^2) this is how it should be.

Answer 1

(3CscX)/(5CscX-6(CotX)^2)
We know 1/sinx = cscx
and cotx = cosx/sinx

So your equation breaks down into:

3(1/sinx)/[5/sinx - 6 (cosx/sinx)^2]

Which becomes:

3/[sinx{5/sinx-6(cosx/sinx)^2}]
Distribute the sin and get:

3/(5-6[cosx]^2/sinx)

It doesn't get much prettier after that though.

Answer 2

csc X = 1/sin X
sec X = 1/cos X
cot X = cos X/sin X
tan X = sin X/cos X

I included all four for thoroughness. It's usually easier to deal just with cosines and sines, so you convert your equation to just include those.

(3/sin X) / ( 5/sinX - 6 cos^2 X/sin^2 X )

You need a common denominator for the fractions in the deonominator, so:

(3/sinX) / (5 sin X/sin^2 X - 6 cos^2 X/sin^2 X
(3/sin X) / [ (5 sin X - 6 cos^2 X)/sin^2 X ]

A sin X cancels out of both the numerator and denominator, leaving you with:

3 / [ (5 sin X - cos^2 X)/sin X ]

The sin X can be moved to the numerator:

3 sin X / [ 5 sin X - cos^2 X ]

cos^2 X = 1 - sin^2 X , so:

3 sin X/ [5 sin X - 1 + sin^2 X] which should be rearranged to:

3 sin X / [ sin^2 X + 5 sin X - 1]

Edit: I didn't catch your squared portion.

Answer 3

first i am not sure how you want this to be expressed (because you did not say in terms of sine or cosine or just csc) but this is what i did

the numerator 3 csc (x) = 3 / sin (x)

the denominator is the tricky part

5 csc (x) - 6 cot ^ 2 (x) can be written as

5 / sin (x) - 6 * [ cos^2 (x) / sin^2 (x)]

now find a common denominator for the denominator
{think sin^2 (x)} and simplify this fraction

[ 5*sin(x) - 6*cos^2(x) ] / sin^2(x)

but we are dividing the numerator 3 / sin (x) by that last expression so the sin(x) will cancel with the sin^2(x) and will leave us with the following

3sin(x) / [ 5*sin(x) - 6*cos^2(x) ]
but i can replace cos^2(x) by 1 - sin^2(x) to get

3sin(x) / [ 5*sin(x) - 6 + 6sin^2(x) ]

now this denominator can be factored (think 6x^2 + 5x - 6)

and you can write it this way

3 sin(x) / [(3 sin(x) - 2)*(2 sin(x) + 3)]

not very pretty and i am not sure if that is what you want but i do hope it helps.

Answer 4

(3CscX)/(5CscX-6(CotX)^2)

looks like trig identity time...

3cscx/ (5cscx - (6cotx)²)

3cscx/ (5cscx - 36 csc²x -36
3/ (5 - 36csc²x - 36) divided cscx out.
3/ (-36csc²x - 31)

3/ (-36 (1/sinx)² -31)
-3/ (36/sin²x + 31)

that's as simple as i can get it

Answer 5

First you opt to get each and each of the instruments to be the same. So the truck is travelling a million mile/minute, so the circumference of the wheel is a million/360 miles, the diameter might want to be a million/360/pi miles. Convert miles to inches 5280*12/360/pi = fifty six.0 inches. A biggggg tire.

Answer 6

(3cscx)/(5csc(x) - 6cot(x)^2)

cot(x)^2 = csc(x)^2 - 1

(3cscx)/(5cscx - 6(csc(x)^2 - 1))

(3csc(x))/(-6csc(x)^2 + 5csc(x) + 6)

(3csc(x))/(-(6csc(x)^2 - 5csc(x) - 6))

ANS : (-3csc(x))/(6(csc(x))^2 - 5csc(x) - 6)

Answer 7

express (3scs(x))/(5csc(x)-6(cot(X)squ...
Change
csc - 1/sinx, scs = 1/cos x and cot x =i/tanx = cos x/sinx
{3/ cos (x)}/ {5/sin (x)} - 6[ cos(x)/ sin(x)]^2
{3/ cos (x)}/ {5/sin (x)} - 6[ cos^2(x)/ sin^2(x)]
{3/ cos (x)}/ {5sin(x) - 6 cos^2(x)}/sin^2x
{3sin^2 (x)/ cos (x)}/ {5sin(x) - 6 cos^2(x)}
3tan (x) sin (x)}/ {5sin(x) - 6 cos^2(x)}