Question

HELP (cot(x)^2)/(csc(x)-1)

Answer 1

\[(\cot (x)^{2})/(\csc(x)-1)\]

Answer 2

\[\frac{ \cot(x)^{2} }{ \csc(x)-1 }\] PLEASE SOLVE WILL GIVE POINTS

Answer 3

I WILL GIVE U A MEDAL I SWEAR

Answer 4

k lets get started

Answer 5

convert the expression into sines and cosines
I hope you know cot(x) = cos(x) /sin(x)
and cosec(x) = 1/sin(x)

Answer 6

Let me know what you get!

Answer 7

(cos(x)^2)/(sin(x)^2) so then i did that and changed csc(x)-1 to cot(x) and then I got cos(x)^3/sin(x) in the end

Answer 8

oh no ...........tell again what is the relation ship b/n csc(x) and cot(x)??

Answer 9

csc(x)-1=cot(x) right?

Answer 10

Can you tell me what identities I use to solve?

Answer 11

no...........the relation is csc^2(x) - 1 = cot^2(x) :P

Answer 12

Then what do I do?

Answer 13

Can you supply the solution?

Answer 14

see u can do the question in a variety of ways.................the best way to do such questions that I found is to convert everything into sine and cos and use a single relation: cos^2x +sin^2x =1

Answer 15

Let us convert everything to sine and cosine

Answer 16

\[ \frac{ \cot^2x }{ \csc x-1 }\]
\[\frac{ \cos^2x/\sin^2x }{ 1/sinx -1 }\]

Answer 17

right??

Answer 18

let me drop you a hint after simplfying the above expression you can write cos^2(x) as 1- sin^2(x) which can again be broken down to (1-sin x)(1+sin x)

Answer 19

I dont know how to simplify further.

Answer 20

I got (sin(x) - sin(x)^3)(1-sin(x)) divided by sin(x)^2

Answer 21

\[(\sin(x)-\sin(x)^3-\sin(x)^2+\sin(x)^4)/(\sin(x)^2)\] this is what i got.

Answer 22

Let me show you

Answer 23

Simplifying from where I wrote earlier
\[\frac{ \cos^2 x. \sin x }{ \sin^2x(1-sin x) }\]

Answer 24

cancel out the sin x and write cos^2x as (1-sin^2x)

Answer 25

you will get
\[\frac{ (1-\sin^2x) }{ \sin x (1-\sin x) }\]

Answer 26

and then i expand (1-sin^2x as)
\frac{ (1-sin x)(1+sin x) }{ sin x (1-sin x) }

Answer 27

I hope you can finish off the rest

Answer 28

Yeah, I really don't get what you did. Anyone care to explain?

Answer 29

hey buddy...
here's an easy method
\[\cot ^{2}x + 1 = \csc ^{2}x\]
Sooooo
\[\cot ^{2}x = \csc ^{2}x -1\]

Now substitute that value in to ur expression
\[\frac{ \cot ^{2}x}{ csc x - 1 } = \frac{ \csc ^{2} x -1 }{ csc x - 1 }\]

Also... u can write
\[\csc ^{2}x -1 =( \csc x+1)(\csc x - 1)\]
now substitute it again to ur expression....
\[\frac{ \csc ^{2} x -1 }{ \csc x - 1 }\ = \frac{ (\csc x -1)(\csc x +1) }{ (\csc x-1) }\]
now cancel out .....
\[(\csc x -1)\] from both numerator and denominator
And ur final answer will be...\[\csc x +1\]

Answer 30

Did ya get it bro ...?

Answer 31

How did csc(x) become squared?

Answer 32

OH i get it thank you!

Answer 33

u r welcome!!