(1+sin x /sin x)=(cot^2x /cscx-1)?

Question

prove or verify

Answer 1

cot^2 x / (csc x - 1)
= (cos^2 x / sin^2 x) / (1/sin x - 1)
= cos^2 x / (sin x - sin^2 x)
= (1 - sin^2 x) / (sin x (1 - sin x))
= (1 + sin x) (1 - sin x) / (sin x (1 - sin x))
= (1 + sin x) / sin x.
(which can be written more nicely as 1 + 1/sin x).

Answer 2

i discover the wonderful component to do as quickly because it truly is truly useful to have the means to desire to memorize multiple formula is to make some flashcards. indoors in ordinary words some training, you will actual memorize 15-20 formula. I as precise now as used that technique to memorize over one hundred thirty formula for an actuary examination. yet yet another trick: take the derivative of a few of your purposes. you're able to be waiting to calculate the derivative of tanx with the aid of capacity of potential of translating to sinx and cosx and applying the quotient rule. one extra suitable trick: use your graphing calc. on your earnings. as quickly as you have a TI-89 you're able to no longer have any subject concerns as that calculator can compute optimal integrals explicitly. yet, as quickly as you have a TI-80 3 or TI-80 4 you will prepare numerical severe to analyze your techniques. i discover it spectacular what sort pupils do no longer use their graphing calculators to their finished earnings. I even have have been given had many ideal exams, with the aid of capacity of actuality I checked my artwork so heavily with a graphing calc. very very final trick: as quickly as you're given a chain of tanx's and secx's you're able to constantly convert those to sinx and cosx, and likewise you're able to hold close what the severe of sinx and cosx are, and be waiting to teach a alleviation formula if necessary.

Answer 3

RHS = (cot^2x / cscx -1)
= [(cos^2x/sin^2x) / (1/sinx-1)]
= [(cos^2x/sin^2x) / (1-sinx/sinx)]
On cancellin sinx we get
= [cos^2x / sinx(1-sinx)]
Multiplying and dividing (1+sinx)
= [cos^2x (1+sinx) / sinx (1-sinx)(1+sinx)]
= [cos^2x (1+sinx) / sinx (1-sin^2x)]
= [cos^2x (1+sinx) / sinx cos^2x)]
Cancelling cos^2x we get
= (1+sinx/sinx)
= LHS