A

Adithya

2xy(x^2-y^2) .,..... Think so

2

A

Aaditya

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)

1

A

Abhijith

ðŸ¤”ðŸ¤”ðŸ¤”

0