Q13/119) Prove that : If .... Tan (x-y)/2 , Tan z and Tan (x+y)/2 are in Geometrical Progression then .cos x = cos y . cos 2 z

we have

tan A tan B + 1 = sin A/cos A sin B/cos B + 1 = ( sin A sin B + cos A cos B)/( cos A cos B)

= cos ( A + B)/ ( cos A cos B) ... (1)

tan A tan B - 1= sin A/cos A sin B/cos B - 1 = ( sin A sin B - cos A cos B)/( cos A cos B)
= - cos ( A - B)/ ( cos A cos B) ... (2)

from (1) and (2) (tan A tan B+ 1)/( tan A tan B- 1) = - cos ( A + B)/cos (A-B)
putting A= (x+y)/2, B= (x-y)/2 we get

(tan (x+y)/2 tan (x-y)/2 + 1)/tan (x+y)/2 tan (x-y)/2- 1) = - cos x / cos y ...(3)

from given condition

tan^2 z = Tan (x-y)2Tan (x+y)/2

or sin ^2 z/ cos ^2 z = Tan (x-y)2Tan (x+y)/2

using componendo dvidendo we get

( sin ^2 z + cos^2 z)/( sin ^2 z - cos^2 z) = (Tan (x-y)2Tan (x+y)/2 +1)/ (Tan (x-y)2Tan (x+y)/2-1)
or 1/(-cos 2z) = - cos x/ cos y

or cos x = cos y cos 2z