Question

verify

sint/1-cost=csct+cott

so far I got the RHS to equal

1+cost/sint

how can I make both sides equal without crossing over?

Answer 1

you're basically done.. :)

Answer 2

\(\large \frac{1+cost}{sint} =\frac{1}{sint} + \frac{cost}{sint} = \) ????

Answer 3

\[\frac{ 1 }{ sint }= csct \]

Answer 4

than

Answer 5

\[\frac{ cost }{ sint }=cott\]

Answer 6

well in my case, when i am stuck on RHS, i see LHS and see what i need whether i can find some connection

Answer 7

what do you think you'll need to do in order to make \[(1+cosx)/sinx\] into\[sinx/(1-cosx)\]

Answer 8

what equation has both sinx and cosx in it?

Answer 9

\[\sin ^{2}t+\cos ^{2}t=1\]

Answer 10

well even i don't know how i got to this..
i'll just show you\[(1+cosx)/sinx = sinx(1+cosx)/\sin^2x = sinx(1+cosx)/(1-\cos^2x)\]

Answer 11

\[\frac{ sint }{ cost }=tant\]

Answer 12

you know how to do the rest right?

Answer 13

let me right it out

Answer 14

what kind of book has these difficult trigonometry questions?

Answer 15

swokowski algebra and trig

Answer 16

all these questions could be on our test tomorrow

Answer 17

i see. good luck:)

Answer 18

thx i need it

Answer 19

so you multiplied the right side with sinx/sinx?

Answer 20

sorry but im stuck, not sure how you got to where you got. where did the sinx come from?

Answer 21

\[\frac{\sin t}{1-\cos t} = \csc t+\cot t\]

Because csc t = 1 / sin t, and cot t = cos t / sin t this becomes:

\[\frac{\sin t}{1-\cos t} = \frac{1+\cos t}{\sin t}\]

Multipling left side by sin t/sin t and right side by (1- cos t)/(1- cos t)
\[\frac{\sin^{2}t}{\sin t\left(1-\cos t\right)} = \frac{\left(1+\cos t\right)\left(1-\cos t\right)}{\sin t\left(1-\cos t\right)}\]

Expanding brackets gives
\[\frac{\sin^2{t}}{\sin t\left(1-\cos t\right)} = \frac{1-\cos^{2}t}{\sin t\left(1-\cos t\right)}\]

We know that sin^2 t + cos^2 t = 1, so sin^2t = 1 - cos^2t:
\[\frac{1-\cos^{2}t}{\sin t\left(1-\cos t\right)} = \frac{1-\cos^{2}t}{\sin t\left(1-\cos t\right)}\]

Answer 22

thanks a million!

Answer 23

sorry i'm late. Meepi is right. again sinx/sinx =1, an identity.

Answer 24

thanks