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Find cot60 degrees-sin45degrees
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\[\cot 60-\sin 45=\frac{\cos 60}{\sin 60}-\frac{\sqrt2}2=\frac{\frac12}{\frac{\sqrt3}2}-\frac{\sqrt2}2=\frac1{\sqrt3}-\frac{\sqrt2}2=\frac{2-\sqrt6}{2\sqrt3}\]
Its suppose to be one of these answers a.)thats for responding but the 4 answers are a)2-sqrt3/2 b.)2 square root3-3 square root 2/6 c)2-squareroot2/2 d)2 square root 2-3 square root 3/6
OK ! We can complete from the last solution : \[\cot 60-\sin 45=\frac{2-\sqrt6}{2\sqrt3}=\frac{2\sqrt3-3\sqrt2}{6}\]
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