Tolong jawabanya beserta caranya,, sin 15°- cos 15° / tan 15° =.... Question from @hidaayanti

October 2018 1 253 Report

Tolong jawabanya beserta caranya,,
sin 15°- cos 15° / tan 15° =.....

izharulhaq78 $\frac{\sin15-\cos15}{\tan15}\\\frac{\sin15-\cos15}{\frac{\sin15}{\cos15}}\\(\sin15-\cos15)(\frac{\cos15}{\sin15})\\\cos15-\frac{\cos^215}{\sin15}\\\cos(45-30)-\frac{(\cos(45-30))^2}{\sin(45-30)}\\(\cos45\cos30-\sin45\sin30)-\frac{(\cos45\cos30-\sin45\sin30)^2}{\sin45\cos30-\cos45\sin30}$
$(\frac{1}{4}\sqrt6+\frac{1}{4}\sqrt2)-\frac{(\frac{1}{4}\sqrt6+\frac{1}{4}\sqrt2)^2}{\frac{1}{4}\sqrt6-\frac{1}{4}\sqrt2}\\(\frac{\sqrt6+\sqrt2}{4})-\frac{(\frac{\sqrt6+\sqrt2}{4})^2}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{\frac{6+2}{16}}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{\frac{8}{16}}{\frac{\sqrt6-\sqrt2}{4}}\\\frac{\sqrt6+\sqrt2}{4}-\frac{1}{2}\times\frac{4}{\sqrt6-\sqrt2}\\\frac{\sqrt6+\sqrt2}{4}-\frac{2}{\sqrt6-\sqrt2}$
$\frac{((\sqrt6)^2-(\sqrt2)^2)-8}{4(\sqrt6-\sqrt2)}\\\frac{(6-2)-8}{4(\sqrt6-\sqrt2)}\\\frac{4-8}{4(\sqrt6-\sqrt2)}\\\frac{-4}{4(\sqrt6-\sqrt2)}\\-\frac{1}{\sqrt6-\sqrt2}\times\frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\\-\frac{\sqrt6+\sqrt2}{6-2}\\\boxed{-\frac{\sqrt6+\sqrt2}{4}}$

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