cosecx / cotx + tanx = cosx
these trig stuff is difficult for me ): in addition to the solution, please add some basic tips and tricks/shortcuts to solving identites.
2007-09-04 22:49:05 · 6 answers · asked by Kendra 3 in Science & Mathematics ➔ Mathematics
no that's the right equation, im certain its 100% correct.
2007-09-04 23:05:00 · update #1
Gilarh er..i don't think that relates to the question above...
2007-09-04 23:07:36 · update #2
thx Rackbrane!!! yes i guess you're right...it takes alot of practice ): & knowing the formulae...it's just that there's so many ways of solving it, gets a bit tedious >:(
2007-09-04 23:18:50 · update #3
The left hand side of your identity needs brackets round the denominator when written on a single line.
cosec(x) / ( cot(x) + tan(x) )
= (1 / sin(x)) / (cos(x)/sin(x) + sin(x)/cos(x))
Putting the denominator over a common denominator sin(x)cos(x) gives:
(cos^2(x) + sin^2(x)) / (sin(x)cos(x))
= 1 / sin(x)cos(x).
Recombining this with the numerator 1 / sin(x) gives:
(1 / sin(x) ) * sin(x)cos(x)
= cos(x).
I don't know any tricks or shortcuts, except to learn the formulae well and practise a lot. What works easily for one question may not work for another.
2007-09-04 23:08:42 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I find the best thing to do when you have to memorize a lot of formulas is to make some flashcards. Within just a few sessions, one can easily memorize 15-20 formulas. I once used that method to memorize over 130 formulas for an actuary exam. Another trick: take the derivative of some of your functions. You should be able to calculate the derivative of tanx by translating to sinx and cosx and applying the quotient rule. One more trick: use your graphing calc. to your advantage. If you have a TI-89 you shouldn't have any problems as that calculator can compute most integrals explicitly. But, if you have a TI-83 or TI-84 you can still use numerical integral to check your answers. I find it amazing how many students don't use their graphing calculators to their full advantage. I've had many perfect tests, because I checked my work so carefully with a graphing calc. Final trick: when you're given a bunch of tanx's and secx's you can always convert these to sinx and cosx, and you should know what the integral of sinx and cosx are, and be able to use a reduction formula if needed.
2016-05-17 07:00:07 · answer #2 · answered by marya 3 · 0⤊ 0⤋
cosecx / cotx + tanx = cosx
cosec x = 1/ sen x
cot x = 1 / tan x
1/sen x / 1/tan x + tan x = cos x
tan x/ senx + tan x = cos x
tan x (1+ sen x)= cos x sen x
senx (1+senx) = cos²x senx
1+senx= cos²x
1-cos²x +senx=0
sen²x+senx=0
if senx=a
a²+a= 0
a(a+1)=0
a1=0
a2=-1
With a1
senx=0
x=0°
With a2
senx=-1
x=270°= 3pi/2
2007-09-04 23:04:57 · answer #3 · answered by Gilarh 3 · 0⤊ 0⤋
The question has been presented incorrectly.
It should read as:-
cosec x / (cot x + tan x) which of course is not what has been given.
cosec x / (cot x + tan x)
(1 / sin x) / (cos x / sin x + sin x / cos x)
Multiply top and bottom by sin x cos x:-
cos x / ( cos ² x + sin ² x)
cos x
PS Remember your brackets!!!
2007-09-05 01:19:14 · answer #4 · answered by Como 7 · 1⤊ 0⤋
csc x / cot x + tan x = cos x
1/sin x sin x / cos x + tan x = cos x
1/ cos x + tan x = cos x
1/ cos x + sin x / cos x = cos x
(1 + sin x) / cos x = cos x
there is something wrong with your equation
2007-09-04 23:03:41 · answer #5 · answered by CPUcate 6 · 0⤊ 1⤋
cosecx/cotx+tanx
=cosecx/cosx/sinx+sinx/cosx[cot=cosx/sinx and tanx=sinx/cosx]
=cosecx/cosx[square]+sinx[square]/sinx into cosx
{but sinx[square]+cosx[square]=1}
=1/sinx /1/sinx cosx
=1/sinx into sinx cosx
=cosx
hence proved
2007-09-04 23:26:14 · answer #6 · answered by whitepaint 2 · 0⤊ 0⤋