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simplify tan x cos x

Mathematics
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Remember tangent is just sin(x)/cos(x)
so its cos x (sin x / cos x) ?
Correct, now what happens when you go ahead and multiply the cosine?

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Other answers:

sin x
That's it. The cosines cancel.
okay cool what bout sec y sin (pi/2-y)? IDK how to do this at all
\[\sec(y)\sin(\frac{\pi}{2-y})\] ??
but isnt it like pi/2 - y not pi/ 2-y
sin(pi/2-y)=cos y sec y= 1/cos y
Ok, so rewrite secant as 1/cos
Then you can use a fancy pants trig identity.
cos y 1/cos y
|dw:1347861835303:dw| Right?
i think so i was kinda looking at what hartnns post too so couldn't you do it like|dw:1347861948541:dw|
sec y =1/ cos y is in numerator, so it would be :|dw:1347862109049:dw|
No, well what you would have is a multiplication, the cosines would cancel.
oh hey you know how tan x= sinx/ cos x well what are the rest of those cuz it might help me understand better like how we getting cos y when its sin y
ok, so u want all trignometric formulas ? for converting one ratio to other ?
yes please
i will list some : \(\huge sin x =1/cosec x \implies cosec x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\ \huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x\)
if u want sin x when cos x is given or if u want cos x when sin x is given, u use: \(sin^2x+cos^2x=1\)
okay
so if its 1+tan^2x/csc^2x that would simplify to sec^2x/csc^2
yes. or is it ? \(\large1+\frac{tan^2x}{cosec^2x}\quad?\)
if it is \(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)
but you put that cos^2x=1+cot^2x
no the last one is cosec^2 x =1+cot^2 x not cos^2x=1+cot^2x
csc=cosec
oh yeah huh cosec is csc not cos i forgot for a minute lol
so is it 1+ tan^2x/csc^2x or sec^2x/csc^2 and why
what exactly is your question ? is this : (1+ tan^2x)/csc^2x
(1+ tan^2x)=sec^2 x from the formula i gave u.
no i saying is it my answer or yours and why (idk how to put it on my computer like you did so that was the best i could do)
to tell u what is your answer, i need to know your exact question, that is why i asked whether it is: (1+ tan^2x)/csc^2x
oh yeah it's that and i need to simplify it
ok,for that i have shown u the steps, tell me which step, u didn't understand: \(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)
oh i get it now earlier you put it wierd so i didn't get it
okay now this one is cot x tan x so it would be (cosx/sinx)(sinx/cosx) correct?
yes, or u could use cot x = 1/ tanx and u get tan x / tan x =1
wait because i have two of each would thay cancel out?
yes, they would. sin x cos x / sin x cos x =1
oh okay just checking
|dw:1347864987907:dw|
okk.
how bout cot u sin u would that simplify to (cosu/sinu)sin u= cos u?
thats absolutely correct! u are improving :)
hooray improvement!
what next ?
1-cos^2 beta/sin beta and that would simplify to ooh okay i got nothin for this one :(
is the numerator (1-cos^2 beta ) ??
yes
from sin^2 beta +cos^2 beta =1 u have 1-cos^2 beta = sin^2 beta so u have : \(\large \frac{sin^2 \beta}{sin \beta}=\frac{sin \beta.sin\beta}{sin \beta}=?\)
sin beta?
yup, but did u get all steps ?
yeah so is that another function 1-cos^2 beta = sin^2 beta
see the list i have given. it has sin^2 x + cos^2 x =1 isn't it ? from here i got 1-cos^2 x = sin^2 x clear ?
oh so you just went backwards?
whenever u see sin^2 x or cos^2 x try to use that formula for simplification.
okay now i have cos x- cos^3 x
like would that be cosx- (cosx)(cosx)/cosx?
nopes. take cos x common from both the terms, what do u get ?
cosx- (cosx)(cosx)(cosx)
so you get 1
no! after taking cos x common, u get cos x (1-(cos x)(cos x) ) =cos x (1- cos^2 x) and u already know what is (1- cos^2 x)
sin^2x
yes! so your final answer will be cos x sin^2x but did u get all the steps ?
yeah because you have coxx-(cosx)(cosx)(cosx) =cosx-1(cosx)=1-cos^2x=sin^2x
i will draw it :|dw:1347866550463:dw|
oh okay then now my last question is sin^2 u + tan^2 u +cos^2 u/sec u
is this entire thing (sin^2 u + tan^2 u +cos^2 u) in the numerator ?
yes
again using sin^2 x +cos^2 x =1, u have numerator as (1+tan^2 x) got this ? and (1+tan^2 x) = sec^2 x.....see formula list. so u have now sec^2x/sec x = ??
replace all of my 'x' with 'u'
oh yeah so now you have sec x!
yup :)
sec u ;)
oh yeah lol thanks for your help i going to sleep now so goodnight^-^
a very good night to u and have sweet dreams :)

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