
#1
Oct805, 12:05 PM

P: 786

Can someone please help me establish this identity?
[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex] 



#2
Oct805, 12:19 PM

P: 93

So, are you proving this identity?
Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a wellknown trigonometric identity.. Can you continue from here? :D 



#3
Oct805, 12:23 PM

P: 786

Thanks! 



#4
Oct805, 04:54 PM

P: 786

Trig Identity
I can't get this one either:
[tex]\frac{1 + \tan \theta}{1  \tan \theta} = \frac{\cot \theta + 1}{\cot \theta  1}[/tex] I'm so bad at proofs 



#5
Oct805, 04:58 PM

HW Helper
P: 661

For this one, you can either choose to replace tan x by 1/cot x or replace cot x by 1/tan x. Choose either and do some algebriac manipulations while leaving the other side alone.




#6
Oct805, 04:59 PM

HW Helper
P: 1,024

Or, if that doesn't work for you, substitute tan by sin/cos and cot by cos/sin, then simplify the expressions
Try, if you get stuck, show us! 



#7
Oct805, 05:16 PM

P: 786

I end up with
[tex]\frac{\cos^2 \theta + \sin \theta \cos \theta}{\cos^2 \theta  \sin \theta \cos \theta}[/tex] or [tex]\frac{\cot^2 \theta + \cot \theta}{\cot^2 \theta  \cot \theta}[/tex] How do I continue? 



#8
Oct805, 05:22 PM

HW Helper
P: 1,024

How did you end up with that?
For the LHS: [tex]\frac{{1 + \tan \theta }}{{1  \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1  \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta  \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta  \sin \theta }}[/tex] Now try the RHS 



#9
Oct805, 05:30 PM

P: 786

Don't I need to continue with the LHS until I get the right or vice versa? 



#10
Oct805, 05:31 PM

HW Helper
P: 1,024

Well now you have the LHS, the easiest would be trying to get the same starting with the RHS, which will go more or less the same




#11
Oct805, 05:34 PM

P: 786

Ah, I see. Thank you both of you.




#12
Oct805, 05:35 PM

HW Helper
P: 1,024

No problem



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