division method of 213,2485 by h
Answers
Step-by-step explanation:
Step-by-step explanation:
we are given that,
secA=17/8
so ,by Pythagoras theorem ,we get,
AB=8k, BC=15k, AC= 17k
SinA= Opp/hypo=BC/AC=15/17
CosA=Adj/hypo=AB/AC=8/17
TanA=Opp/Adj=15/8
Now,
3-4sin^2A/4cos^A-3 = 3-Tan^A/1-3Tan^A
3-4(15/17)^2/4(8/17)^2-3 = 3-(15/8)^2/1-3(15/8)^2
3-4(225/289)/4(64/289)-3 = 3-(225/64)/1-3(225/64)
3-(900/289)/(256/289)-3 = 3-(225/64)/1-(675/64)
(867-900)/289/(256-867)/289 = (192-225)/64/(64-675)/64
(-33)/298/(-611)/289 = (-33)/64/(-611)/64
(-33)/(-611) = (-33)/(-611)
33/611 = 33/611
hence,proved
LHS=RHS
Answer:
Step-by-step explanation:
we are given that,
secA=17/8
so ,by Pythagoras theorem ,we get,
AB=8k, BC=15k, AC= 17k
SinA= Opp/hypo=BC/AC=15/17
CosA=Adj/hypo=AB/AC=8/17
TanA=Opp/Adj=15/8
Now,
3-4sin^2A/4cos^A-3 = 3-Tan^A/1-3Tan^A
3-4(15/17)^2/4(8/17)^2-3 = 3-(15/8)^2/1-3(15/8)^2
3-4(225/289)/4(64/289)-3 = 3-(225/64)/1-3(225/64)
3-(900/289)/(256/289)-3 = 3-(225/64)/1-(675/64)
(867-900)/289/(256-867)/289 = (192-225)/64/(64-675)/64
(-33)/298/(-611)/289 = (-33)/64/(-611)/64
(-33)/(-611) = (-33)/(-611)
33/611 = 33/611
hence,proved
LHS=RHS