Question

7th baar puchh chukapar koi achhe se nhi batata

solve​

Answer 1

$\huge\sf\purple{Solution}$

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

LHS=RHS

Hence, proved.

Answer 2

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