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So LHS becomes :
\[LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}\]

yea i got that too... then i prob stuffed up somewhere...

Got it..

never seen this before :P
\[\huge\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}\]

I too... :P

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???

yea :P

yes

i know them :)

Just divide them.. :)

You will get formula for \(tan^2(u)\).

Now I think you can prove it.. Go ahead..

ahh yes i can :) thank you :)

You are welcome dear..