Question

cot(45-A).cot(45-B)=1

Answer 1

To do what?

Answer 2

prove.. this question

Answer 3

prove it...or find solutions.?? if solutions r needed..thn we have cot (45-A).cot(45-B)=1...so..cot(45-A)=tan(45-B)...right...now remember tan A=cot(90-A)..so we rewrite this as tan(90-(45-A))=tan(45-B)...so we have..tan(45+A)=tan(45-B)...so you can find the relation betn A nd B..using general solutions..

Answer 4

trigo gen solns u'll get in any standrd maths book..

Answer 5

how cot(45-A)=tan(45-B)??

Answer 6

be fast!!

Answer 7

remember..1/tan x=cot x nd vice versa...so cot(45-A).cot(45-B)=1...that implies that cot(45-A)=(1/cot(45-B))...thats cot(45-A)=tan(45-b)..getting it..??

Answer 8

nopes.. :-(

Answer 9

u do get this that cot(45-A)=(1/cot(45-B))...don't u..??

Answer 10

got it... i proved it... thnks!!

Answer 11

:)

Answer 12

now next question!!! i had posted .... r8 now

Answer 13

i got no ques...in the forum...can u pste it here..??

Answer 14

btw which class r u in now??

Answer 15

11th...

Answer 16

okkkk...so ur ques..paste it here...plz..

Answer 17

cot(45-A).cot(45+A)=1..prove it

Answer 18

okkk...so use the same property here...cot(45-A)=tan(90-(45-A))...lyk i told u in the last ques..so cot(45-A)=tan(45+A)...so our expression becomes..tan(45+A)*cot(45+A)...so lyk I told u...cotx*tanx is always 1...so proved..

Answer 19

got it??

Answer 20

yup... now next.. :-) if u dnt mind then...

Answer 21

no but i deserve a medal..:P :P

Answer 22

i didnt know how to give someone medal.. in this site!! :-p

Answer 23

lol..no probs...type ur ques here..but...along with typing in the forum..in case i fail..

Answer 24

okk..

Answer 25

cot2x.cotx-cot2xcot3x-cot3xcotx=1

Answer 26

okk..m on it...ths will tke tym..

Answer 27

hmm.....

Answer 28

cot(a+b) = (cosa cosb - sina sinb) / (sina cosb + sinb cosa)
=cosa cosb / (sina cosb + sinb cosa) - sina sinb / (sina cosb + sinb cosa)
= 1/ (sina/cosa + sinb/cosb) - 1/(cosb/sinb + cosa/sina)
= 1/ (tana + tanb) - 1/ (cota + cotb)
So cot(3x) = 1/(tanx + tan2x) - 1/(cotx + cot2x)
so
cotx.cot2x - cot2x.cot3x - cot3x.cotx
=cotx.cot2x - (cot2x+cotx).cot3x
=cotx.cot2x - (cot2x+cotx).(1/(tanx + tan2x) - 1/(cotx + cot2x))
=cotx.cot2x - (cotx+cot2x)/(tanx+tan2x) + 1
Put m = cotx, and n=cot2x
= mn - (m+n)/(1/m + 1/n) + 1
= mn - (m+n)/(n/mn + m/mn)) + 1
= mn - (m+n)/((m+n)/mn) + 1
= mn - mn + 1
= 1
get it...??...ws a long one..i used some formulae that can be derieved from simple ones..

Answer 29

wait let me try!!

Answer 30

okkk...

Answer 31

got it/... next!!!

Answer 32

yeah okay...hey u preparing for iit??...whr r u frm??

Answer 33

\[\tan2\theta+ \tan3\theta-\tan5\theta=-\tan2\theta.\tan3\theta.\tan5\]

Answer 34

5theta rite?..not tan 5..

Answer 35

yup... by mistake.. i had wrtten

Answer 36

tell me in simply way.. plzzz

Answer 37

this will probabbly use tan(a+b+c)...so i'm not so sure it'll be very simple...although m working on it..nd m nt sure..

Answer 38

nopes its wriitn that the formila cot(A+B)or cot(A-B)...

Answer 39

yeah my mistake...look tan 5A=tan(2A+3A)....remember the formula for tan(x+y)?..tan(x+y)=(tan x +tan y)/(1-tanx.tany)...so now..tan 5A=tan(2A+3A)..thts ( tan 2A + tan 3A)/(1-tan 2A.tan 3A) thn tan 5A(1-tan 3A tan 2A) = tan 3A +tan 2A
tan 5A -tan 5A tan 3A tan 2A = tan 3A +tan 2A
tan 5A - tan 3A - tan 2A = tan 5A.tan 3A tan 2A..got it??...sory i wnt to the toilet..so lt rply

Answer 40

let me try!! :)

Answer 41

:)yeah...hey but r u indian..??

Answer 42

yup.... ad u??

Answer 43

me too...from whr r u..???

Answer 44

next ... cotA.cotB+cotB.cotC+cotC.cotA=1

Answer 45

if A+B+C= pie...

Answer 46

yes..tht was d conditn...

Answer 47

ok...A+B=180-C...rite??...so..tan(A+B)=tan(180-C)...now tan(180-C)=-tan C remember...??...so
(tan A + tan B)/(1-tan A tan B) = - tan C

or tan A + tan B + tan C = tan A tan B tan C

divide both sides by tan A tan B tan C...this is very common process in mny sums..

so 1/tan B 1/ tan C + 1/tan A 1/ tan C + 1/ tan A 1/tan B = 1

or cot B cot C+ cot A cot C + cot A cot B = 1

Answer 48

got it??

Answer 49

yup.....

Answer 50

tan18+tan27 tan18tan27=1

Answer 51

should dere be a+ u missd..chck the ques..

Answer 52

nopes...

Answer 53

tan 27.tan 18. tan27 shud be (tan27)^2.tan18...rite??

Answer 54

not got....

Answer 55

didn't get me..??? u wrote tan18+tan 27tan 18tan 27=1...so tan 27 tan 18tan27 shud be (tan 27)^2tan18..rite???...did u mean dat...?

Answer 56

tan18+tan27 +tan18tan27=1

Answer 57

the quest is this...

Answer 58

yupppp....now it can be done..

Answer 59

so...tan 18+tan27=1-tan 18tan 27..rite...so...(tan 18+tan 27)/(1-tan 18.tan 27)=1...but see...dats the formula for tan(x+y)...lyk (tan x+tan y)/(1-tan xtan y)...correct??...so u getn tan (27+18)=1...thats tan 45=1...which is true offcourse ...got it??

Answer 60

yup....

Answer 61

so next prob?

Answer 62

\[\sqrt{3}(\tan140-\tan110)= 1+\tan 140.\tan110\]

Answer 63

hey...this one is easy...lyk the last one...
just rearrange it...
(tan 140-tan 110)/(1+tan 140.tan 110)=(1/(3)^1/3)
thats the form of tan (x-y) remember...?...so tan (140-110)=(1/(3)^1/3)...thats tan 30=(1/(3)^1/3)...thats true...

Answer 64

wats this????

Answer 65

the proof...??u didn't get it??

Answer 66

na

Answer 67

ok..so 1.44(tan 140-tan 110)=1+tan 140.tan 110...so we can write (tan 140-tan 110)/(1+tan 140tan 110)=1/1.44=.69...see the left hand side is of the form tan (x-y) thats (tan x-tan y)/(1+tan x.tan y)...so we can write tan (140-110)=.69...thats tan 30=.69...which is true...u get it now...Btw 1.44 is cube root of 3...

Answer 68

hey u got this??

Answer 69

no.. leave it...

Answer 70

how come u r not getting this...its easy blv me...tan(x-y)=(tan x-tan y)/(1+tanx.tany)...u have to rearrange the terms to get this form...see if u can get that...it wud be betr...if i could actually show u on pen nd paper..