If `tantheta+tan(theta+pi/3)+tan(theta-pi/3)=ktan3theta` then k is equal to

tan theta+tan(theta+(pi)/(3))+tan(theta-(pi)/(3))=k tan36 then k is equal to

tan 3theta = tan 2theta + tantheta

Solve tan theta+tan(theta+pi/3)+tan (theta+2pi/3)=3

If theta_1,theta_2,theta_3 are the three values of thetain[0,2pi] for which tantheta=lambda then the value of tan(theta_1/3) tan(theta_2/3) + tan(theta_2/3) tan(theta_3/3) + tan(theta_3/3) tan(theta_1/3) is equal to

find the value of k which satisfy tan theta +tan(theta+pi/3)+tan (theta-pi/3)=ktan 3 theta

sin^(3)theta = tan2theta + tantheta

If tan theta + tan (pi/3 +theta) + tan (frac{2pi}{3}+ theta) = k tan 3theta then k is equal to

sqrt(3)tantheta tan(theta + pi/3) tan(theta + (2pi)/3) =1