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Mathematics 21 Online
OpenStudy (anonymous):

prove .

OpenStudy (anonymous):

|dw:1346666692924:dw|

OpenStudy (anonymous):

change tan into sin/cos everywhere

OpenStudy (anonymous):

and cot too ?

OpenStudy (anonymous):

then u,ll have cos3A.CosA/sin2A + sin3A.SinA/Sin2A

OpenStudy (anonymous):

dont do ath to RHS

OpenStudy (anonymous):

okay !

OpenStudy (anonymous):

after simplifying last expression u.ll get cos2A/Sin2A...which is RHS

OpenStudy (anonymous):

i didn't understand the last step !

OpenStudy (anonymous):

u got this expression....cos3A.CosA/sin2A + sin3A.SinA/Sin2A???

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok then after taking LCM in numerator u have...cos3A.CosA+sin3A.SinA...which is equal to Cos(3A-A) i.e. cos2A

OpenStudy (anonymous):

use d identity Cos(A-B)= cosAcosB+ SinA.SinB

OpenStudy (anonymous):

okay , got it now .. thank you @akash123 :)

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