Question

If tanx= sina.sinb/cosa+cosb then prove that tan x/2 = tan a/2.tanb/2

Answer 1

Answer:

Step-by-step explanation:

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

Answer:

tan(x)/2=tan(a)/2.tan(b)/2