# The value of  (cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )??

Aug 10, 2018

$\left(\cos \left(\frac{\pi}{12}\right) - \sin \left(\frac{\pi}{12}\right)\right) \left(\tan \left(\frac{\pi}{12}\right) + \cos \left(\frac{\pi}{12}\right)\right)$

$= \left(\cos \frac{\frac{\pi}{12}}{\cos} \left(\frac{\pi}{12}\right) - \sin \frac{\frac{\pi}{12}}{\cos} \left(\frac{\pi}{12}\right)\right) \cos \left(\frac{\pi}{12}\right) \left(\tan \left(\frac{\pi}{12}\right) + \cos \left(\frac{\pi}{12}\right)\right)$

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

=(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))

=(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))

Now $\tan \left(\frac{\pi}{12}\right) = \tan \left(\frac{\pi}{3} - \frac{\pi}{4}\right)$

$= \frac{\tan \left(\frac{\pi}{3}\right) - \tan \left(\frac{\pi}{4}\right)}{1 + \tan \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{4}\right)} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Again
$\sin \left(\frac{\pi}{12}\right)$

$= \sin \left(\frac{\pi}{3} - \frac{\pi}{4}\right)$

$= \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

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

$= \left(1 - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) \left(\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{2} \left(1 + \frac{\sqrt{3}}{2}\right)\right)$

$= \left(\frac{2}{\sqrt{3} + 1}\right) \left(\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{8} \left(4 + 2 \sqrt{3}\right)\right)$

$= \left(\sqrt{3} - 1\right) \left(\frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{1}{8} {\left(\sqrt{3} + 1\right)}^{2}\right)$

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

$= \left({\left(\sqrt{3} - 1\right)}^{2} / \left(2 \sqrt{2}\right) + \frac{1}{4} \left(\sqrt{3} + 1\right)\right)$

$= \left(\frac{4 - 2 \sqrt{3}}{2 \sqrt{2}} + \frac{1}{4} \left(\sqrt{3} + 1\right)\right)$

$= \left(\frac{2 - \sqrt{3}}{\sqrt{2}} + \frac{1}{4} \left(\sqrt{3} + 1\right)\right)$

$= \left(\frac{2 \sqrt{2} - \sqrt{6}}{2} + \frac{1}{4} \left(\sqrt{3} + 1\right)\right)$