cos2a/seca-sin2a/coseca=cos3a
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Answered by
14
easy one
(Cos2A/SecA)-(Sin2A/CosecA) = Cos3A
from LHS,
(Cos2A/SecA)-(Sin2A/CosecA)
= {Cos2A/(1/CosA)}-{Sin2A/(1/SinA)}
= Cos2ACosA-SinASin2A
= Cos (2A+A)
= Cos3A=RHS
hence proved
(Cos2A/SecA)-(Sin2A/CosecA) = Cos3A
from LHS,
(Cos2A/SecA)-(Sin2A/CosecA)
= {Cos2A/(1/CosA)}-{Sin2A/(1/SinA)}
= Cos2ACosA-SinASin2A
= Cos (2A+A)
= Cos3A=RHS
hence proved
poojith51:
what about sinasin2a
Formula of Cos (A+B) = CosACosB-SinASinB ...SinA cames from CosecA =1/SinA and CosA cames from SecA= 1/CosA
ok bro thanks
Answered by
5
L.H.S
=(Cos2A/SecA)-(Sin2A/CosecA)
=(Cos2A/1/CosA)-(Sin2A/1/SinA)
=Cos2A*CosA-Sin2A*SinA
=Cos(2A+A)
=Cos3A. R.H.S. Hence proved
=(Cos2A/SecA)-(Sin2A/CosecA)
=(Cos2A/1/CosA)-(Sin2A/1/SinA)
=Cos2A*CosA-Sin2A*SinA
=Cos(2A+A)
=Cos3A. R.H.S. Hence proved
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