Suggested: tan pi/2 - tan(pi/2-x) - tan inverse x-1/x-2+tan inverse x+1/x+2=pi/4 - cosx/1-sinx=tan(pi/4+x/2) - prove that tan(pi/4+x)/tan(pi/4-x)=(1+tanx/1-tanx)^2 - u=log tan(pi/4+theta/2) - if u=log tan((pi)/(4)+(theta)/(2)) then coshu= - sin^(2)(pi)/(6)+cos^(2)(pi)/(3)-tan^(2)(pi)/(4)=-(1)/(2) - 1-sin2x/1+sin2x=tan^2(pi/4-x) - limit of tan x as x approaches pi/2 - log tan (pi/4+x/2) - tan inverse sin minus pi by 2 - tan inverse tan 2 pi by 3 - tan(pi 2) Browse related:
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