At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

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?

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?

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.

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

yes please

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.

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

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

sin beta?

yup, but did u get all steps ?

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

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

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

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 :)