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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
 11 months ago
 11 months ago
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
 11 months ago
 11 months ago

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

waterineyesBest ResponseYou've already chosen the best response.1
So LHS becomes : \[LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}\]
 11 months ago

AonZBest ResponseYou've already chosen the best response.0
yea i got that too... then i prob stuffed up somewhere...
 11 months ago

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

waterineyesBest 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...
 11 months ago

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

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

waterineyesBest ResponseYou've already chosen the best response.1
So, moving forward: \[\frac{1 + \cos(4A)}{\sqrt{1 + \cos(4A)} \sqrt{1  \cos(4A)}} \implies \frac{\sqrt{1 + \cos(4A)}}{{\sqrt{1  \cos(4A)}}}\]
 11 months ago

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

waterineyesBest ResponseYou've already chosen the best response.1
Now use the 1 formula: \[\implies \frac{1}{\sqrt{\tan^2(2A)}} \implies \frac{1}{\tan(2A)}\]
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
Use the second formula here and you will get your answer...
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
Oh, I forget, you need verification of that formula???
 11 months ago

waterineyesBest 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.. :(
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
Just divide them.. :)
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
You will get formula for \(tan^2(u)\).
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\huge \frac{\sin^2(u)}{\cos^2(u)} = \frac{\frac{1  \cos(2u)}{2}}{\frac{1 + \cos(2u)}{2}}\] \[\tan^2(u) = \frac{1 \cos(2u)}{1 + \cos(2u)}\]
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
Now I think you can prove it.. Go ahead..
 11 months ago

AonZBest ResponseYou've already chosen the best response.0
ahh yes i can :) thank you :)
 11 months ago

waterineyesBest ResponseYou've already chosen the best response.1
You are welcome dear..
 11 months ago
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