Prove that:`"cot"pi/(24)=sqrt(2)+sqrt(3)+sqrt(4)+sqrt(6)`

1.	Prove that:`"cot"pi/(24)=sqrt(2)+sqrt(3)+sqrt(4)+sqrt(6)`
Answer» `cot (pi/24) = cos (pi/24)/sin(pi/24)` `=cos (pi/24)/sin(pi/24)(2cos(pi/24))/(2cos(pi/24))` `=(2cos^2 (pi/24))/(2sin(pi/24)cos(pi/24))` `=(1+cos(pi/12))/(sin(pi/12))` `=(1+cos(pi/4-pi/6))/(sin(pi/4-pi/6))` `=(1+cos(pi/4)cos(pi/6)+sin(pi/4)sin(pi/6))/(sin(pi/4)cos(pi/6)-cos(pi/4)sin(pi/6))` `=(1+(1/sqrt2)(sqrt3/2)+(1/sqrt2)(1/2))/((1/sqrt2)(sqrt3/2)-(1/sqrt2)(1/2))` `=(2sqrt2+sqrt3+1)/(sqrt3-1)` `=(2sqrt2+sqrt3+1)/(sqrt3-1)(sqrt3+1)/(sqrt3+1)` `=(2sqrt6+3+sqrt3+2sqrt2+sqrt3+1)/(3-1)` `=sqrt6+sqrt2+sqrt3+2` `=sqrt2+sqrt3+sqrt4+sqrt6 = R.H.S.`

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