If a, b are positive quantities and if a1=a+b2,b1=a1b,a2=a1+b12,b2=a2b1 and so on, then

Let a = bcosθ,then a1=a+b2=b(1+cos⁡θ)2=bcos2⁡θ2⇒b1=a1b=bcos⁡θ2Now, a2=a1+b12=bcos⁡θ2cos2⁡θ4∴b2=bcos⁡θ2cos⁡θ22Similarly, b3=bcos⁡θ2cos⁡θ22cos⁡θ23and so on.Now, bn=bcos⁡θ2cos⁡θ22…cos⁡θ2n∴b∞=limn→∞ bsin⁡θ2nsin⁡θ2n=limn→∞ bsin⁡θθsin⁡θ/2nθ/2n=bsin⁡θθ=b1−cos2⁡θcos−1⁡(a/b)∴b∞=b2−a2cos−1⁡(a/b)