Question

factoring and express this into tg(alpha) terms:

tg(alpha)sec^2 (alpha) + ctg(alpha)csc^2 (alpha)

thanks

Answer 1

tg? ctg?

Answer 2

tg(alpha)

Answer 3

what? that doesn't answer my question...
tg(alpha) = tan(alpha) or something?

Answer 4

I assume tg is tangent and ctg is cotangent... so why not use tan and cot like a normal person :P

Answer 5

I don't live in USA. Most of spanish math books about trygonometry express this functions like that

tan (alpha) = tg(alpha)

It's the same with its opposite function...

Answer 6

I need help because I got an answer, however it doesn't match with my math book result ... So I just want to check your procedures, if you can of course... thanks

Answer 7

\[\large \tan \alpha \sec^2 \alpha + \cot \alpha \csc^2 \alpha\]use some identities \\[\large \tan \alpha (1 +\tan^2 \alpha) \alpha + \frac{ 1 }{ \tan \alpha } \left( 1+ \frac{ 1 }{ \tan^2 \alpha } \right)\]from there just simplify it

Answer 8

You need to use identities.

Answer 9

\[[1 + \tan^2(\alpha)][\tan^2(\alpha) + \frac{ 1 }{ \tan^3(\alpha) }]\] this is the math book answer, however I don't get this...

Could you get this?

Answer 10

What an odd way of writing it :/\[ \large \tan \alpha (1 +\tan^2 \alpha) + \frac{ 1 }{ \tan \alpha } \left( 1+ \frac{ 1 }{ \tan^2 \alpha } \right)\] \[ \large \tan \alpha +\tan^3 \alpha+ \frac{ 1 }{ \tan \alpha } + \frac{ 1 }{ \tan^3 \alpha } \]

Answer 11

It's OK, I got that you have written but, math book result is the last expression I put before... My question is

Could you get the expression I put before? I mean

\[[1 + \tan^2(\alpha)] [\tan^2(\alpha) + \frac{ 1 }{ \tan^3(x) }] \]

Could you get this exactly?

Answer 12

starting from your last expression

\[\tan(\alpha) + \tan^3(\alpha) + \frac{ 1 }{ \tan(\alpha) } + \frac{ 1 }{ \tan^3(\alpha) }\]

Could you get the one I put before? @agent0smith

Answer 13

\[\large [1 + \tan^2(\alpha)] [\tan^2(\alpha) + \frac{ 1 }{ \tan^3 \alpha }] = \] \[\large \tan^2 \alpha +\tan^4 \alpha +\frac{ 1 }{ \tan \alpha }+\frac{ 1 }{ \tan^3 \alpha }\]does not appear equivalent to what I have above; first two terms are multiplied by tan alpha.

Answer 14

So, the math book result isn't correct! :)

I was confused about my procedure but I just realized I was right! thanks