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AonZ
Group Title
Prove cosec4A + cot4A = 1/2 (cotA tanA)
Any way of doing this easily? like with minimum working out?
@jim_thompson5910
cause i get stuck with a massive page of working out
 one year ago
 one year ago
AonZ Group Title
Prove cosec4A + cot4A = 1/2 (cotA tanA) Any way of doing this easily? like with minimum working out? @jim_thompson5910 cause i get stuck with a massive page of working out
 one year ago
 one year ago

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waterineyes Group TitleBest ResponseYou've already chosen the best response.1
How about using these first: \[cosec(4A) = \frac{1}{\sin(4A)}\] \[\cot(4A) = \frac{\cos(4A)}{\sin(4A)}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
So LHS becomes : \[LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}\]
 one year ago

AonZ Group TitleBest ResponseYou've already chosen the best response.0
yea i got that too... then i prob stuffed up somewhere...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Got it..
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
You want to just remember these formulas here: \[\tan^2(u) = \frac{1  \cos(2u)}{1 + \cos(2u)}\] \[\tan(2u) = \frac{2\tan(u)}{1  \tan^2(u)}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
So in the numerator above there is one 1+ cos(4A) is there but we need there one (1cos(4A), so that we can apply the formula 1 there...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
By using identity \(sin^2(x) + cos^2(x) = 1\) : Sin(4A) can be written as: \[\sin(4A) = \sqrt{1  \cos^2(4A)} \implies \sqrt{1  \cos(4A)} \cdot \sqrt{1 +\cos(4A)}\]
 one year ago

AonZ Group TitleBest ResponseYou've already chosen the best response.0
never seen this before :P \[\huge\tan^2(u) = \frac{1  \cos(2u)}{1 + \cos(2u)}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I too... :P
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
So, moving forward: \[\frac{1 + \cos(4A)}{\sqrt{1 + \cos(4A)} \sqrt{1  \cos(4A)}} \implies \frac{\sqrt{1 + \cos(4A)}}{{\sqrt{1  \cos(4A)}}}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Now just take reciprocal, you will get the same form: \[\huge \frac{1}{\frac{\sqrt{1  \cos(4A)}}{\sqrt{1 + \cos(4A)}}}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Now use the 1 formula: \[\implies \frac{1}{\sqrt{\tan^2(2A)}} \implies \frac{1}{\tan(2A)}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Use the second formula here and you will get your answer...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Oh, I forget, you need verification of that formula???
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
You know this?? \[\sin^2(u) = \frac{1  \cos(2u)}{2}\] \[\cos^2(u) = \frac{1 + \cos(2u)}{2}\] Now don't say you have not seen them too.. :(
 one year ago

AonZ Group TitleBest ResponseYou've already chosen the best response.0
i know them :)
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Just divide them.. :)
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
You will get formula for \(tan^2(u)\).
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \frac{\sin^2(u)}{\cos^2(u)} = \frac{\frac{1  \cos(2u)}{2}}{\frac{1 + \cos(2u)}{2}}\] \[\tan^2(u) = \frac{1 \cos(2u)}{1 + \cos(2u)}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Now I think you can prove it.. Go ahead..
 one year ago

AonZ Group TitleBest ResponseYou've already chosen the best response.0
ahh yes i can :) thank you :)
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
You are welcome dear..
 one year ago
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