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eliminate(thita) from the equations atan(thita)+bcot2(thita)=c and acot(thita)-btan2(thita)=c Badiuddin askIITians.ismu Expert
148 Points
11 years ago

Dear ravi

 atanθ+bcot2θ=c  ........1 acotθ-btan2θ=c  ..........2

subtract equation

a(cotθ -tanθ)=b(cot2θ+tan2θ)

a cos2θ/cosθsinθ = b/cos2θ sin2θ

2a/tan2θ  = 2b/sin4θ

sin4θ /tan2θ =b/a

2tan2θ/(1+tan22θ).tan2θ =b/a

let T =tan2θ

2T/(1+T2)T =b/a

T2 =(2a-b)/b  ....................3

and from the equation 1 and 2

c-b/T =atanθ

c+bT =acotθ

multiply

(c-b/T)(c+bT) =a2

c2 -b2 +bc(T-1/T) =a2

c2 -b2 +bc(T2-1)/T =a2

put the value of T

c2 -b2 +bc(2a-2b)/b  .√b/(√2a-b) =a2

2c(a-b)√b =(a2+b2-c2)√(2a-b)

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