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Mathematics 14 Online
OpenStudy (anonymous):

Find the exact value of the following expression. If undefined, state, undefined. sec((cot^(-1)(5/12)) can some one pls write this down step by step: I ma going nuts trying to use the the identity sec2x=1+tan2x

OpenStudy (anonymous):

do I first change cot to tan? then square it?

OpenStudy (anonymous):

Akileez i explained this earlier on - check it out.

OpenStudy (anonymous):

thanks uzma; I knwo you replied previously to it

OpenStudy (anonymous):

I know jimmyrep; I wen tover it again and was confused

OpenStudy (anonymous):

cot^-1(5/12)=x 5/12=cotx tanx=12/5

OpenStudy (anonymous):

I did< i understand from using a right angles triangle. but i want to understand from a substitution perspective too.

OpenStudy (anonymous):

now use the identity

OpenStudy (anonymous):

oh ok yes change cot to tan tan = 1/cot then square it and plug it into the formula

OpenStudy (anonymous):

ok got that, but what about the sec in front after using the identity?

OpenStudy (anonymous):

take the sqrt

OpenStudy (anonymous):

and how/why can you put an equals sign from tan(12/5) to tanx=12/5

OpenStudy (anonymous):

is it by adding the x component?

OpenStudy (anonymous):

\[secx=\sqrt{1+(12/5)^2}\]

OpenStudy (anonymous):

cotx=5/12 i/tanx=5/12 tanx=12/5

OpenStudy (anonymous):

cot is the reciprocal of tan

OpenStudy (anonymous):

yes, got that

OpenStudy (anonymous):

this is what i get:

OpenStudy (anonymous):

sec(sec^2x=25/144)

OpenStudy (anonymous):

sec(sec^2x=(25/144) + 1)

OpenStudy (anonymous):

how come?

OpenStudy (anonymous):

just solve what i posted earlier

OpenStudy (anonymous):

I'll show you what I did: sec(tan(5/12))

OpenStudy (anonymous):

now I originally changed tan to sin over cos and changed sec to I/cos - but that left me with sin

OpenStudy (anonymous):

my issue is that I do not see how you manage to cancel the trig funtions out and come up with just a number

OpenStudy (anonymous):

I really appreciate your help as this is bothering me as it shld be so obvious,

OpenStudy (anonymous):

u r not need to change it, we have got the value of tanx, plug it into the identity n do the computations

OpenStudy (anonymous):

its ok, just be clear at ur end :)

OpenStudy (anonymous):

will do:

OpenStudy (anonymous):

can we start with the cot^-1 part first ?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

u post what u think

OpenStudy (anonymous):

that would change to tan

OpenStudy (anonymous):

therefore you now have sec(tan(5/12)) - is that true>?

OpenStudy (anonymous):

no its not right

OpenStudy (anonymous):

ok, this is where I am stumped - sorry.

OpenStudy (anonymous):

let cot^-1(5/12)=x 5/12=cotx tanx=12/5

OpenStudy (anonymous):

ok?

OpenStudy (anonymous):

remind that what is x taken to be

OpenStudy (anonymous):

ok, got that - forgot the ^(-1)

OpenStudy (anonymous):

therefore it shld now read sec(tan(12/5))

OpenStudy (anonymous):

is that correct?

OpenStudy (anonymous):

now dealing with the tan part first: now, tan^2x+(12/5)^2 is that right?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

ok, so what goes after sec(tan(12/5)) ?

OpenStudy (anonymous):

or are you first dealing with tanx=12/5?

OpenStudy (anonymous):

how did u get it?

OpenStudy (anonymous):

yes after getting tanx=12/5 use the identity

OpenStudy (anonymous):

sec x

OpenStudy (anonymous):

ok great: then after squaring: I will have sec^2x-1=144/25

OpenStudy (anonymous):

= (144/25) - 1

OpenStudy (anonymous):

sec^2x=1+144/25

OpenStudy (anonymous):

i have an idea. draw a 5 - 12 - 13 right triangle. cotangent is "adjacent over opposite" so label the angle accordingly. that is label the angle whose opposite adjacent side is 5 and whose opposite side is 12. then all the other trig functions of that angle should be obvious

OpenStudy (anonymous):

( sorry plus one) then I will have to solve: sec(1 +144/25)

OpenStudy (anonymous):

yes sat.. u r right, but he/she wanted to do with identity :)

OpenStudy (anonymous):

sec x=13/5

OpenStudy (anonymous):

thanks anyway Sat

OpenStudy (anonymous):

so you dont solve it further?

OpenStudy (anonymous):

it is done , y further?

OpenStudy (anonymous):

what I mean is:

OpenStudy (anonymous):

ok do it the hard way, fine with me!

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

you have sec(169/25) right?

OpenStudy (anonymous):

haha; sorry, hard way it is

OpenStudy (anonymous):

no no,

OpenStudy (anonymous):

actually it became hardest :P

OpenStudy (anonymous):

from sec(144/25+1)

OpenStudy (anonymous):

all the identities come right out of the triangle anyway. i will be quiet now

OpenStudy (anonymous):

haha; i am sooooo sorry: that i am such a pain, just really want to get it

OpenStudy (anonymous):

now satellite is here, he is so sweet, would make u understand

OpenStudy (anonymous):

its ok, no pain, no gain :)

OpenStudy (anonymous):

up to this part: sec^2x=1+144/25

OpenStudy (anonymous):

right

OpenStudy (anonymous):

take squre root

OpenStudy (anonymous):

oh first simplify the right side

OpenStudy (anonymous):

ok, that leaves us with sec 13/5

OpenStudy (anonymous):

sorry; first simplify

OpenStudy (anonymous):

u missed x again

OpenStudy (anonymous):

169/25

OpenStudy (anonymous):

sec^2x =169/25

OpenStudy (anonymous):

so sec^2(x) = 169/25

OpenStudy (anonymous):

sec x=13/5

OpenStudy (anonymous):

therefore sec =13/5

OpenStudy (anonymous):

my issue i show we did away with the sec outside the (cot^(-1)(5/12)) - am I being clear?

OpenStudy (anonymous):

yup :)

OpenStudy (anonymous):

when u get secx =13/5 put the value of x back sec(cot^(-1)(5/12)=13/5

OpenStudy (anonymous):

oh now I see!!!

OpenStudy (anonymous):

yay!!! i am so happy! thank you millions and millions!

OpenStudy (anonymous):

my pleasure, if i made u understand, i m done

OpenStudy (anonymous):

practically it all came dow to saying let cot^(-1)(5/12) = x

OpenStudy (anonymous):

you rock!!!

OpenStudy (anonymous):

hahaha, i asked u keep it in mind

OpenStudy (anonymous):

I was so busy trying to cross out the trig functions, that's why I missed it all the time

OpenStudy (anonymous):

thank you so much for your help!

OpenStudy (anonymous):

take care:)

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