The value of $(cos (\frac{π}{12}) - sin (\frac{π}{12})) (tan (\frac{π}{12}) + cos (\frac{π}{12}))$ $(cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )$ ??

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The value of $(cos (\frac{π}{12}) - sin (\frac{π}{12})) (tan (\frac{π}{12}) + cos (\frac{π}{12}))$ $(cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )$ ??

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Answer 1 · 2018-08-10T14:42:28

$(cos (\frac{π}{12}) - sin (\frac{π}{12})) (tan (\frac{π}{12}) + cos (\frac{π}{12}))$ $(cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )$

$= (\frac{cos (\frac{π}{12})}{cos (\frac{π}{12})} - \frac{sin (\frac{π}{12})}{cos (\frac{π}{12})}) cos (\frac{π}{12}) (tan (\frac{π}{12}) + cos (\frac{π}{12}))$ $= (cos (pi/12)/cos (pi/12)-sin (pi/12)/cos (pi/12))cos (pi/12)(tan (pi/12)+cos( pi/12) )$

$= (1 - tan (\frac{π}{12})) (sin (\frac{π}{12}) + {cos}^{2} (\frac{π}{12}))$ $=(1-tan(pi/12))(sin(pi/12)+cos^2(pi/12))$

$= (1 - tan (\frac{π}{12})) (sin (\frac{π}{12}) + \frac{1}{2} (1 + cos (\frac{π}{6}))$ $=(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))$

Now $tan (\frac{π}{12}) = tan (\frac{π}{3} - \frac{π}{4})$ $tan(pi/12)=tan(pi/3-pi/4)$

$= \frac{tan (\frac{π}{3}) - tan (\frac{π}{4})}{1 + tan (\frac{π}{3}) tan (\frac{π}{4})} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ $=(tan(pi/3)-tan(pi/4))/(1+tan(pi/3)tan(pi/4))=(sqrt3-1)/(sqrt3+1)$
Again
$sin (\frac{π}{12})$ $sin(pi/12)$

$= sin (\frac{π}{3} - \frac{π}{4})$ $=sin(pi/3-pi/4)$

$= sin (\frac{π}{3}) cos (\frac{π}{4}) - cos (\frac{π}{3}) sin (\frac{π}{4})$ $=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)$

$= \frac{\sqrt{3} - 1}{2 \sqrt{2}}$ $=(sqrt3-1)/(2sqrt2)$
So
$(1 - tan (\frac{π}{12})) (sin (\frac{π}{12}) + \frac{1}{2} (1 + cos (\frac{π}{6}))$ $(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))$

$= (1 - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}) (\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{2} (1 + \frac{\sqrt{3}}{2}))$ $=(1-(sqrt3-1)/(sqrt3+1))((sqrt3-1)/(2sqrt2)+1/2(1+sqrt3/2))$

$= (\frac{2}{\sqrt{3} + 1}) (\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{8} (4 + 2 \sqrt{3}))$ $=(2/(sqrt3+1))((sqrt3-1)/(2sqrt2)+1/8(4+2sqrt3))$

$= (\sqrt{3} - 1) (\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{8} {(\sqrt{3} + 1)}^{2})$ $=(sqrt3-1)((sqrt3-1)/(2sqrt2)+1/8(sqrt3+1)^2)$

$= ⎛ ⎜ ⎜ ⎝ \frac{{(\sqrt{3} - 1)}^{2}}{2 \sqrt{2}} + \frac{1}{8} (\sqrt{3} - 1) {(\sqrt{3} + 1)}^{2} ⎞ ⎟ ⎟ ⎠$ $=((sqrt3-1)^2/(2sqrt2)+1/8(sqrt3-1)(sqrt3+1)^2)$

$= ⎛ ⎜ ⎜ ⎝ \frac{{(\sqrt{3} - 1)}^{2}}{2 \sqrt{2}} + \frac{1}{4} (\sqrt{3} + 1) ⎞ ⎟ ⎟ ⎠$ $=((sqrt3-1)^2/(2sqrt2)+1/4(sqrt3+1))$

$= (\frac{4 - 2 \sqrt{3}}{2 \sqrt{2}} + \frac{1}{4} (\sqrt{3} + 1))$ $=((4-2sqrt3)/(2sqrt2)+1/4(sqrt3+1))$

$= (\frac{2 - \sqrt{3}}{\sqrt{2}} + \frac{1}{4} (\sqrt{3} + 1))$ $=((2-sqrt3)/(sqrt2)+1/4(sqrt3+1))$

$= (\frac{2 \sqrt{2} - \sqrt{6}}{2} + \frac{1}{4} (\sqrt{3} + 1))$ $=((2sqrt2-sqrt6)/2+1/4(sqrt3+1))$

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