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```        eliminate(thita) from the equations
atan(thita)+bcot2(thita)=c
and      acot(thita)-btan2(thita)=c```
7 years ago

147 Points
```										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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```
7 years ago
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