Question

cos theta= (cosa-cosb)1- cosacosb then tan theta/2=?

Answer 1

 have considered question as: cos x=(cosA-cosB)/(1-cosAcosB) 
cos x=[1-tan^2(x/2)]/[1+tan^2(x/2)] 
by componendo-dividendo on LHS andRHS 
[(-1)/tan^2(x/2)]= 
[(cosA-cosB+1-cosAcosB)/(cosA-cosB-1+c... 
factorising numerator and denominator of RHS 
[(-1)/tan^2(x/2)]= 
[{(1+cosA)(1-cosB)}/{(cosA-1)(1+cosB)}... 
[{cos^2(A/2)sin^2(B/2)}/{-sin^2(A/2)co... 
{-cot^2(a/2)tan^2(B/2)} 
cancelling negative sign on both sides and taking reciprocal we get required result.

Hope This Helps :)

Answer 2

tanq/2=±tana/2cotb/2