Question

Good afternoon buddies, could you express this into ctg(alpha) terms?

tg^2 (alpha) + tg(alpha) -tg(alpha)*sec^2 (alpha)

thanks

Answer 1

Use this identity for \(\tan \theta\)

\(\tan \alpha = \dfrac{1}{\cot \alpha} \)

We need to take care of \(\sec^2\theta\)

\(1 + \tan^2 \theta = \sec^2 \theta\)

Answer 2

mmmm perfect my friend, I already know that. However my math book answer says that the result is this :

\[\frac{ 4 + 3ctan^2(\alpha) + 2ctan^4(\alpha) }{ctan^2(\alpha)[1 +ctan^2(\alpha)] }\]

I got another different... So, could you get this expression in ctan(alpha) function factoring ?

Answer 3

\[\large \tan^2 \alpha + \tan \alpha -\tan \alpha \sec^2 \alpha\] \[\large \frac{ 1 }{ \cot^2 \alpha } + \frac{ 1 }{ \cot \alpha } - \frac{ 1 }{ \cot \alpha } (1 + \frac{ 1 }{ \cot^2 \alpha })\] \[\large \frac{ 1 }{ \cot^2 \alpha } + \frac{ 1 }{ \cot \alpha } - \frac{ 1 }{ \cot \alpha } + \frac{ 1 }{ \cot^3 \alpha }\] \[\large \frac{ 1 }{ \cot^2 \alpha } + \frac{ 1 }{ \cot^3 \alpha }= \frac{ \cot \alpha + 1 }{ \cot^3 \alpha}\]

Answer 4

@agent0smith your procedure is wrong

Answer 5

there's a little mistake

Answer 6

Makes very litle diff. though\[\large \frac{ 1 }{ \cot^2 \alpha } + \frac{ 1 }{ \cot \alpha } - \frac{ 1 }{ \cot \alpha } - \frac{ 1 }{ \cot^3 \alpha }\] \[\large \frac{ 1 }{ \cot^2 \alpha } - \frac{ 1 }{ \cot^3 \alpha }= \frac{ \cot \alpha - 1 }{ \cot^3 \alpha}\]

Answer 7

that's the answer I got but, math book answer says it's the one I wrote before above

Answer 8

Again your book has demonstrated how much it sucks:
http://www.wolframalpha.com/input/?i=%5Ctan%5E2+%5Calpha+%2B+%5Ctan+%5Calpha+-%5Ctan+%5Calpha+%5Csec%5E2+%5Calpha+%3D+%5Cfrac%7B+4+%2B+3cot%5E2(%5Calpha)+%2B+2cot%5E4(%5Calpha)+%7D%7Bcot%5E2(%5Calpha)%5B1+%2Bcot%5E2(%5Calpha)%5D++%7D

Answer 9

I'm solving it... each one of the trygonometry exercises. I have discovered there are some mistakes

Answer 10

I was gonna try simplifying the book's expression, till i realized the numerator of theirs is prime, so I didn't even bother.