Tan(pi/3)

Limit="lim_ (x to pi/6) (1-tan2x)/ (2tanx), = (1-tan (2pi/6))/ (2tan (pi/6)), = (1-tan (pi/3))/ (2*1/sqrt3), = (1-sqrt3)/ (2/sqrt3 . # (1-tan^2 (theta))/ (1+sin^2 (theta))= (2/cos^2 (theta))-1# let # \ \ \ \ theta=pi/3# # (1-tan^2 (pi/3))/ (1+sin^2 (pi/3))= (2/cos^2 (pi/3))-1# # (-2)/ (7/4)=2/ (1/4)-1# X= (2k-1)pi, kinzz we have, (2sin ( (5pi)/6)+cosx)/ (tan (pi/2)+cos ( (2pi)/3))=0 =>2sin ( (5pi)/6)+cosx=0 =>2sin (pi-pi/6)+cosx=0 =>2sin (pi/6)+cosx=0 =>2xx1/2+cosx .

(df)/dx|_ (x=-pi/3) = -tan (-pi/3) = sqrt3 (11pi)/6 4cos^2 x - 2 = sec^2 x - tan^2 x (1) develop the right side: How do you find the exact values of tan 67.5 degrees using the half angle formula?

How do you find the exact values of tan pi/ 8 using the half angle formula?