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# [Core 3] Trigonometry watch

1. Realised I struggle with this topic quite a bit and probably will continue to do so, so might as well use one thread for it.

First question I'm stuck on:

"By letting tanx=t, show that the equation 4tan2x + 3cotx sec^(2)x = 0 becomes (3t)^4-(8t)^2-3=0"

I split each bit of the first equation into parts to try and get them in terms of t.
4tan2x became 8t/(1-(t)^2) using the double angle formula
3cotx = 3/t as I know cot = 1/tan
Couldn't get sec^2x in terms of t, not sure how to proceed.
2. 1+tan^2x = sec^2x

(Original post by DarkEnergy)
Realised I struggle with this topic quite a bit and probably will continue to do so, so might as well use one thread for it.

First question I'm stuck on:

"By letting tanx=t, show that the equation 4tan2x + 3cotx sec^(2)x = 0 becomes (3t)^4-(8t)^2-3=0"

I split each bit of the first equation into parts to try and get them in terms of t.
4tan2x became 8t/(1-(t)^2) using the double angle formula
3cotx = 3/t as I know cot = 1/tan
Couldn't get sec^2x in terms of t, not sure how to proceed.
3. (Original post by NotNotBatman)
1+tan^2x = sec^2x
Thanks, haven't been taught that.
4. (Original post by DarkEnergy)
Thanks, haven't been taught that.
It comes from

dividing through by cos^2x gives :

It follows,

Try dividing by sin^2x and you'll get another identity.
5. (Original post by NotNotBatman)
It comes from

dividing through by cos^2x gives :

It follows,

Try dividing by sin^2x and you'll get another identity.
cosec^2x=cot^2x+1 ?
6. (Original post by DarkEnergy)
cosec^2x=cot^2x+1 ?
yes.

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