2xy(x^2-y^2) .,..... Think so
Aaditya Kiran K Sujith.,where did u find the JEE monthly contest.??
1+cos(A)=2cos^2(A/2) 1-cos(A)=2sin^2(A/2) Putting these in above qstn. We get, {Cos^2(A/2)}/{Sin^2(A/2)}^1/2 =cotA/2=x/y => tan A/2=y/x also tan A= (2tan A/2) / (1-tan^2A/2) Finally we get ans.= 2xy/(x^2-y^2)