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June 30th, 2019, 02:35 PM  #11 
Newbie Joined: Jan 2017 From: London Posts: 10 Thanks: 0  They are identities. Try it yourself substitute x on the left hand side and on the right hand side and you will get the same angle. The exercise is just to demonstrate this is true by following it through logically and step by step. Last edited by skipjack; July 1st, 2019 at 04:44 AM. 
July 1st, 2019, 04:51 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
I'm questioning that. Had you checked it out in detail (including any implicit assumptions), requiring precise definitions at all stages, you'd have found a flaw.

July 1st, 2019, 09:32 AM  #13  
Newbie Joined: Jan 2017 From: London Posts: 10 Thanks: 0  Quote:
The path that greg1313 posted makes sense to me. Please point out the flaw... Last edited by skipjack; July 1st, 2019 at 12:20 PM.  
July 1st, 2019, 12:18 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
Did greg1313 start with (1 + sin x)/cos x or did you simply assume that was the case?

July 2nd, 2019, 09:53 AM  #15 
Newbie Joined: Jan 2017 From: London Posts: 10 Thanks: 0  
July 2nd, 2019, 10:24 AM  #16  
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573  Quote:
$\dfrac{\cos{x}}{1\sin{x}} \implies \sin{x} \ne 1 \implies x \ne (4k+1) \cdot \dfrac{\pi}{2} \, ; \, k \in \mathbb{Z}$ ... there are values of x where the right side equals 0, whereas the left side never equals 0  

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