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AonZ
 one year 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
AonZ
 one year 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

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

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

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

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

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1So 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...

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

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

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

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

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

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

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

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1You 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.. :(

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

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

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

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

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

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