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|dw:1346666692924:dw|
change tan into sin/cos everywhere
and cot too ?
then u,ll have cos3A.CosA/sin2A + sin3A.SinA/Sin2A
dont do ath to RHS
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okay !
after simplifying last expression u.ll get cos2A/Sin2A...which is RHS
i didn't understand the last step !
u got this expression....cos3A.CosA/sin2A + sin3A.SinA/Sin2A???
yes
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ok then after taking LCM in numerator u have...cos3A.CosA+sin3A.SinA...which is equal to Cos(3A-A) i.e. cos2A
use d identity Cos(A-B)= cosAcosB+ SinA.SinB
okay , got it now .. thank you @akash123 :)
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