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Trigonometry 14 Online
OpenStudy (anonymous):

Verify identity cot(θ)+tan(θ)=csc(θ)sec(θ) Any hints how to start? I'm completely stuck on this one... Thanks.

OpenStudy (tkhunny):

Last Resort: Change everything to sine and cosine.

OpenStudy (anonymous):

ok let me see

OpenStudy (anonymous):

ok so i got (cos(θ)+sin(θ))/(cos(θ)+sin(θ))=(1)/(cos(θ)sin(θ))

OpenStudy (anonymous):

stuck again

OpenStudy (anonymous):

left side will be 1

OpenStudy (anonymous):

what do i do with right side?

OpenStudy (sirm3d):

the addition of fraction is not right. \[\frac{a}{b} + \frac{b}{a} \color{red}\neq \frac{a+b}{b+a}\]

OpenStudy (anonymous):

you right... ok, let me redo it, thx

OpenStudy (anonymous):

ok, so when i did multiply it as it should be I got (1)/(cos(θ)sin(θ)) on both sides. is that correct?

OpenStudy (anonymous):

\[\frac{ \cos \theta }{ \sin \theta } + \frac{ \sin \theta }{ \cos \theta } \] then multiply by the common factors so first fraction by cos theta and the second by sin theta so you get : \[\frac{ \cos^2 \theta + \sin^2 \theta }{ \sin \theta \cos \theta}\] according to the pythagorean the numenator will equal 1 so you get 1/ sin theta cos theta

OpenStudy (anonymous):

thats what i got, perfect thx

OpenStudy (anonymous):

on the other side csc theta sec theta can be simplified to: (1/sin theta ) (1/cos theta) so you get 1 / sin theta cos theta QED

OpenStudy (anonymous):

QED and the trigie biggie

OpenStudy (anonymous):

EXACTLY

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