Suggested: tan pi/4 - tan pi/4+x/tan pi/4-x - tan((pi)/(4)+(x)/(2))=sec x+tan x - cosx/1-sinx=tan(pi/4+x/2) - prove that tan(pi/4+x)/tan(pi/4-x)=(1+tanx/1-tanx)^2 - cos2x/1+sin2x=tan(pi/4-x) - u=log tan(pi/4+theta/2) - tan inverse tan 3 pi by 4 - expand tan(x+pi/4) using taylor series - expand tan inverse x in powers of x-pi/4 - tan inverse x-1/x-2+tan inverse x+1/x+2=pi/4 - tan-1(2x)+tan-1(3x)=pi/4 - 1-sin2x/1+sin2x=tan^2(pi/4-x) - sin^(2)(pi)/(6)+cos^(2)(pi)/(3)-tan^(2)(pi)/(4)=-(1)/(2) - tan pi 4 Browse related:
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