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Let 2tan `√[a-b/a+b].tan∆/2=xTan x/2=√a-b/a+b.tan∆/2 We know that;Cos x=1-tan^ x/2/1+tan^ x/2Now substitute tan x value in above equation;Cos x=1-(a-b /a+b )tan^∆/2//1+(a-b /a+b ).tan^x/2Now take Lcm.a+b-(a-b)tan^∆/2//a+b+ (a-b) tan^∆/2now substitute tan^∆/2=sin^∆/2//cos^∆/2a(cos^∆/2-sin^∆/2) +b(cos^∆/2+sin^∆/2)//a(cos^∆/2+sin∆/2)+b(cos^∆/2-sin^∆/2)We know that cos^∆/2-sin^∆/2=cos∆and cos^∆/2+sin^∆/2=1Therefore cos x=acos∆+b//a+bcos∆;X=cos`[acos∆+b//a+bcos∆]So option.A