- anonymous

simplify tan x cos x

- chestercat

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- anonymous

Remember tangent is just sin(x)/cos(x)

- anonymous

so its cos x (sin x / cos x) ?

- anonymous

Correct, now what happens when you go ahead and multiply the cosine?

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## More answers

- anonymous

sin x

- anonymous

That's it. The cosines cancel.

- anonymous

okay cool
what bout sec y sin (pi/2-y)? IDK how to do this at all

- anonymous

\[\sec(y)\sin(\frac{\pi}{2-y})\]
??

- anonymous

but isnt it like pi/2 - y not pi/ 2-y

- hartnn

sin(pi/2-y)=cos y
sec y= 1/cos y

- anonymous

Ok, so rewrite secant as 1/cos

- anonymous

Then you can use a fancy pants trig identity.

- anonymous

cos y 1/cos y

- anonymous

|dw:1347861835303:dw|
Right?

- anonymous

i think so i was kinda looking at what hartnns post too so couldn't you do it like|dw:1347861948541:dw|

- hartnn

sec y =1/ cos y is in numerator, so it would be :|dw:1347862109049:dw|

- anonymous

No, well what you would have is a multiplication, the cosines would cancel.

- anonymous

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

- hartnn

ok, so u want all trignometric formulas ? for converting one ratio to other ?

- anonymous

yes please

- hartnn

i will list some :
\(\huge sin x =1/cosec x \implies cosec x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\
\huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x\)

- hartnn

if u want sin x when cos x is given or if u want cos x when sin x is given, u use:
\(sin^2x+cos^2x=1\)

- anonymous

okay

- anonymous

so if its 1+tan^2x/csc^2x that would simplify to sec^2x/csc^2

- hartnn

yes.
or is it ?
\(\large1+\frac{tan^2x}{cosec^2x}\quad?\)

- hartnn

if it is
\(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)

- anonymous

but you put that cos^2x=1+cot^2x

- hartnn

no the last one is cosec^2 x =1+cot^2 x
not cos^2x=1+cot^2x

- hartnn

csc=cosec

- anonymous

oh yeah huh cosec is csc not cos i forgot for a minute lol

- anonymous

so is it 1+ tan^2x/csc^2x or sec^2x/csc^2 and why

- hartnn

what exactly is your question ?
is this : (1+ tan^2x)/csc^2x

- hartnn

(1+ tan^2x)=sec^2 x from the formula i gave u.

- anonymous

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)

- hartnn

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

- anonymous

oh yeah it's that and i need to simplify it

- hartnn

ok,for that i have shown u the steps, tell me which step, u didn't understand:
\(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)

- anonymous

oh i get it now earlier you put it wierd so i didn't get it

- anonymous

okay now this one is cot x tan x so it would be (cosx/sinx)(sinx/cosx) correct?

- hartnn

yes, or u could use cot x = 1/ tanx and u get tan x / tan x =1

- anonymous

wait because i have two of each would thay cancel out?

- hartnn

yes, they would.
sin x cos x / sin x cos x =1

- anonymous

oh okay just checking

- hartnn

|dw:1347864987907:dw|

- hartnn

okk.

- anonymous

how bout cot u sin u would that simplify to
(cosu/sinu)sin u= cos u?

- hartnn

thats absolutely correct!
u are improving :)

- anonymous

hooray improvement!

- hartnn

what next ?

- anonymous

1-cos^2 beta/sin beta and that would simplify to ooh okay i got nothin for this one :(

- hartnn

is the numerator (1-cos^2 beta ) ??

- anonymous

yes

- hartnn

from
sin^2 beta +cos^2 beta =1
u have
1-cos^2 beta = sin^2 beta
so u have :
\(\large \frac{sin^2 \beta}{sin \beta}=\frac{sin \beta.sin\beta}{sin \beta}=?\)

- anonymous

sin beta?

- hartnn

yup, but did u get all steps ?

- anonymous

yeah
so is that another function 1-cos^2 beta = sin^2 beta

- hartnn

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 ?

- anonymous

oh so you just went backwards?

- hartnn

whenever u see sin^2 x or cos^2 x
try to use that formula for simplification.

- anonymous

okay now i have cos x- cos^3 x

- anonymous

like would that be cosx- (cosx)(cosx)/cosx?

- hartnn

nopes.
take cos x common from both the terms, what do u get ?

- hartnn

cosx- (cosx)(cosx)(cosx)

- anonymous

so you get 1

- hartnn

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)

- anonymous

sin^2x

- hartnn

yes!
so your final answer will be
cos x sin^2x
but did u get all the steps ?

- anonymous

yeah because you have coxx-(cosx)(cosx)(cosx) =cosx-1(cosx)=1-cos^2x=sin^2x

- hartnn

i will draw it :|dw:1347866550463:dw|

- anonymous

oh okay then now my last question is sin^2 u + tan^2 u +cos^2 u/sec u

- hartnn

is this entire thing (sin^2 u + tan^2 u +cos^2 u) in the numerator ?

- anonymous

yes

- hartnn

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 = ??

- hartnn

replace all of my 'x' with 'u'

- anonymous

oh yeah so now you have sec x!

- hartnn

yup :)

- hartnn

sec u ;)

- anonymous

oh yeah lol thanks for your help i going to sleep now so goodnight^-^

- hartnn

a very good night to u and have sweet dreams :)

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