Show that tan(π/6+ A/2).tan (π/6- A/2)=(2 cos A-1)/(2 cos A+1)

Hi Varun,

From the identities as u know are given below:
tan(A-B) = sin(A-B)/cos(A -B) .................................(1)
tan(A+B) =sin(A+B)/cos(A +B)  .................................(2)

Multiplying the eqns. (1) and (2) we get,
tan(A-B) * tan(A+B) = 2[sin(A-B)sin(A+B)]/2[cos(A – B)*cos(A+B)]
: – ) {Multiplying from 2 in numerator denominator both sides}

R.H.S becomes = = =>      (cos2B – cos2A)/(cos2B + cos2A).....................(3)
Now put here according to the requireed questiom,

A = pie/6  and B = A/2
Dont correlate A and B they are independent of each other,,.
ust put the values in eqn. (3)
U will get further, 

(cosA – ½) / (cosA +1/2) = same as Rh.s after taking L.C.M

hOPE it helps!