Question

(cosx/tanx)(1-sec^2(x))

Answer 1

Another one with trig identities?

Answer 2

I am really bad a these lol

Answer 3

This is another on that uses the fact that you gotta simplify tan first and now we're gonna use another new one: \[\Huge 1+\cot^2(x)=\sec^2(x)\]

Answer 4

so 1-sec^2(x) would become 1-1+cot^2(x)--> cot^(2)x?

Answer 5

No sorry.

Answer 6

\[\Huge 1+\tan^2(x)=\sec^2(x)\]

Answer 7

so 1-1+tan^2(x)--> tan^2x?

Answer 8

Okay, so I plugged in what I just got and my end result is sinx, is that right?

Answer 9

I found on a website that tan^2(x)+1=sec^2(x), does the position of the "1" matter?

Answer 10

Yes

Answer 11

Well on the one side of the equation, no

Answer 12

Like you can have 1+tan^2(x)=sec^2(x) or tan^2(x)+1=sec^2(x) but you can't have somehting like tan^2(x)=sec^2(x)+1 or tan^2(x)=1+sec^2(x),

Answer 13

because then you'd be moving the one to another side of the equation when you should really subtract it on both sides so yeah you'd have tan^2(x)=sec^2(x)-1

Answer 14

so this is how I arranged it: 1-sec^2(x)--> 1-(tan^2(x)+1)-->(1-tan^2(x)-1)--->-tan^2(x). then I did (cos(x)/tan(x)) * (-tan^2(x)) --->cos(x) * -tan^2(x)---> cos(x)* (sin^2(x))/cos^2(x)) --> -sinx^2(x)/cosx that's what i have so far, am I on the right track?

Answer 15

What is this equal to? Or are you just supposed to simplify it?

Answer 16

simplify!

Answer 17

I got a new result of -sin(x)

Answer 18

is that okay?

Answer 19

\[\frac{\cos x}{\frac{\sin x}{\cos x}}(1-\frac{1}{\cos ^2x})=\frac{\cos ^2x}{\sin x}(1-\frac{1}{\cos ^2x})\]

Answer 20

is that your final result?

Answer 21

no

Answer 22

\[\frac{\cos ^2x}{\sin x}(\frac{\cos ^2x-1}{\cos ^2x})=\frac{-\sin ^2x}{\sin x}=-\sin x\]

Answer 23

is it alright if I did 1-sec^2(x)--> 1-(tan^2(x)+1)?

Answer 24

yes

Answer 25

thank you!!

Answer 26

yw