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Mathematics 24 Online
OpenStudy (anonymous):

Simplify the trig expression. cos(x)(sec(x)-cos(x))

OpenStudy (anonymous):

start with \[\cos(x)\left(\frac{1}{\cos(x)}-\cos(x)\right)\]

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i still have no idea where to go.

OpenStudy (unklerhaukus):

expand the brackets!

OpenStudy (anonymous):

use the distributive property to multiply then use the most famous trig identity of them all

OpenStudy (anonymous):

i got 1-cos^2x

OpenStudy (unklerhaukus):

do you remember the most famous trig identity?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

nvm is it sin^2x???

OpenStudy (unklerhaukus):

something^2+something^2 = 1

OpenStudy (anonymous):

yeah sin^2x+cos^2x=1

OpenStudy (unklerhaukus):

cool, can you rearrange this to be cos^2x = ....

OpenStudy (anonymous):

=1-sin^2x

OpenStudy (unklerhaukus):

you had 1-cos^2x now sub in cos^2x = 1-sin^2x

OpenStudy (anonymous):

wait what?

OpenStudy (anonymous):

im getting sin^2x

OpenStudy (unklerhaukus):

Yep, that's right

OpenStudy (anonymous):

oh ok cool

OpenStudy (anonymous):

what about (tanx+cotx) cotx

OpenStudy (unklerhaukus):

use cot = 1 / tan

OpenStudy (anonymous):

im actually splittting the expression into 2 fractions. tanx/cotx + cotx/cotx

OpenStudy (anonymous):

thats what i did first anyway

OpenStudy (anonymous):

then cot over cot is just 1.

OpenStudy (anonymous):

so (tanx/cotx)+1

OpenStudy (anonymous):

then what u said cot=1/tan... so my result is = tan^2x+1 which is sec^2x

OpenStudy (unklerhaukus):

\[\checkmark\]

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