Prove that the given equations are identities: 1+tan(theta)tan2(theta)=tan2(theta)cot(theta)-1

Question

lgbasallote · Answer

do you know the equation for $\tan 2\theta$? coz i have forgotten and i might be able to help if i know it

rath111 · Answer

Yes I do know. Hold on...

rath111 · Answer

tan2(theta)=2tan(theta)/1-tan^2(theta)

lgbasallote · Answer

gimme a minute to figure this out...

rath111 · Answer

Okay.

lgbasallote · Answer

have you tried it? if so how far did you get?

rath111 · Answer

I've tried it but... I got to 1+tan(theta)tan2(theta)=tan^2(theta)/1-tan^2(theta)

Sure I did something wrong though...

lgbasallote · Answer

hmm you did the right hand side huh....i got sec^2 theta/1 - tan^2 theta for that

rath111 · Answer

Someone was typing a reply, then he went silent.

lgbasallote · Answer

$\LARGE \frac {2 \tan \theta \cot \theta}{1 - \tan^2 \theta} - 1$

$\LARGE \frac {2 (\frac{\cancel{\sin \theta}}{\cancel{\cos \theta}}) (\frac{\cancel{\cos \theta}}{\cancel{\sin \theta}})}{1 - \tan^2 \theta} - 1$

$\LARGE \frac {2 }{1 - \tan^2 \theta} - 1$

$\LARGE \frac {2 }{1 - \tan^2 \theta} - 1$

lol this is what i got...ignore the sec something i said a while ago

rath111 · Answer

I don't know... What? I'm confused.

anonymous · Answer

1+tan(theta)tan2(theta)=tan2(theta)cot(theta)-1 ok so lets do right side first and change everything to sin and cos like we did before okay so tan2(theta)=2tan(theta)/1-tan^2(theta) and cot theta = costheta/sintheta ok so i am going to change theta into x's alright

tan2xcotx - 1 = so this is just (sin2x/cos2x)(cosx/sinx) - 1 = (2sinxcosx/cos^2x - sin^2x)(cosx/sinx) - 1 = (2cos^2x/cos^2x - sin^2x) - 1 = (2cos^2x - cos^2x +sin^2x)/cos^2x - sin^2x the top just becomes a 1 so now we got 1/cos2x for the right side okay :)

anonymous · Answer

now for the left side 1+tan(theta)tan2(theta) change everything to sin and cos alright so we got 1+ (tanx)(tan2x) and that is 1+ (sinx/cosx)[(2sinxcosx/(cos^2x - sin^2x)] = 1 + [2sin^x/(cos^2x - sin^2x)] = [(cos^2x - sin^2x + 2sin^2x)/(cos^2x - sin^2x)] the top just becomes a 1 so you got 1/cos2x just like we wanted :) alright @Rath111

paxpolaris · Answer

$$RHS=\LARGE \frac {2 }{1 - \tan^2 \theta} - 1={2-1+\tan^2(\theta) \over 1-\tan^2(\theta)}$$

then do the same with LHS

lgbasallote · Answer

@PaxPolaris correct me if im wrong but in proving i think you are only allowed to touch one side

paxpolaris · Answer

no you can simplify both sides, if you need to.

paxpolaris · Answer

you could also first re-write the original equation into: $$\Large 2= \tan(2\theta)\left[ \cot(\theta)- \tan(\theta)\right]$$

rath111 · Answer

Trying to absorb this information...

rath111 · Answer

The left side is what's so trippy right now...

rath111 · Answer

k getting it...

rath111 · Answer

That took forever for my weak brain to process, but I finally understood it after messing with the problem and looking back and forth with what I thought was the best answer. Thank you :)