Question

I'm stuck on these two questions in Precalc. Please help? I have the answers, but I would like to know how to get them.

1. (tan^3x-1)/(tanx-1)-sec^2x-1=0, cotx=? [answer is 1]

2. x=1/2tanx where -pi/2 12 years ago

Answer 1

first problem in equation editor ..?? plz
and what do have to find in ques.2

Answer 2

\[(\tan^3\theta-1/\tan \theta -1)-\sec^2\theta -1=0\]

Answer 3

For the second one, I'm supposed to write f(x) in terms of a single trigonometric function of theta.

Answer 4

\[\frac{ (tanx-1)(\tan^2x+1+tanx )}{ tanx-1} -(1+\tan^2x)-1\]

Answer 5

to make that fast i used x instead of theta ...so as u can see ..apply the identity of \[a^3-b^3\]

Answer 6

so cotx =1 after simplifying

Answer 7

That makes sense, thanks.

Answer 8

in the second one ...does\[x=\frac{ 1 }{ 2tanx }\]

Answer 9

\[x=1/2(tanx)\] is that the same thing?

Answer 10

because by that my answer is coming as 1/2secx

Answer 11

\[f(x)=\frac{ \frac{ 1 }{ 2tanx } }{ \sqrt{1+\frac{ 4 }{ 4\tan^2x }} }\]

Answer 12

\[f(x)=\frac{ \frac{ 1 }{ 2tanx} }{ \frac{ secx }{ tanx } }\]

Answer 13

hope u r getting that .??

Answer 14

Yes, I think so.

Answer 15

so hence ..the answer is 1/2secx ..??

Answer 16

Maybe the top of the equation should be 2/tanx? x is supposed to be (1/2)(tanx). Does that make it come out to 1/2sinx?

Answer 17

yes then it comes that .!

Answer 18

Okay, I think I've got it now :) Thanks for your help!