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AonZ
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
How about using these first: \[cosec(4A) = \frac{1}{\sin(4A)}\] \[\cot(4A) = \frac{\cos(4A)}{\sin(4A)}\]
So LHS becomes : \[LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}\]
yea i got that too... then i prob stuffed up somewhere...
You want to just remember these formulas here: \[\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}\] \[\tan(2u) = \frac{2\tan(u)}{1 - \tan^2(u)}\]
So in the numerator above there is one 1+ cos(4A) is there but we need there one (1-cos(4A), so that we can apply the formula 1 there...
By using identity \(sin^2(x) + cos^2(x) = 1\) : Sin(4A) can be written as: \[\sin(4A) = \sqrt{1 - \cos^2(4A)} \implies \sqrt{1 - \cos(4A)} \cdot \sqrt{1 +\cos(4A)}\]
never seen this before :P \[\huge\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}\]
So, moving forward: \[\frac{1 + \cos(4A)}{\sqrt{1 + \cos(4A)} \sqrt{1 - \cos(4A)}} \implies \frac{\sqrt{1 + \cos(4A)}}{{\sqrt{1 - \cos(4A)}}}\]
Now just take reciprocal, you will get the same form: \[\huge \frac{1}{\frac{\sqrt{1 - \cos(4A)}}{\sqrt{1 + \cos(4A)}}}\]
Now use the 1 formula: \[\implies \frac{1}{\sqrt{\tan^2(2A)}} \implies \frac{1}{\tan(2A)}\]
Use the second formula here and you will get your answer...
Oh, I forget, you need verification of that formula???
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.. :(
Just divide them.. :)
You will get formula for \(tan^2(u)\).
\[\huge \frac{\sin^2(u)}{\cos^2(u)} = \frac{\frac{1 - \cos(2u)}{2}}{\frac{1 + \cos(2u)}{2}}\] \[\tan^2(u) = \frac{1- \cos(2u)}{1 + \cos(2u)}\]
Now I think you can prove it.. Go ahead..
ahh yes i can :) thank you :)
You are welcome dear..