Statement Find the exact values (then approximate to four decimals) of: (a) sec150∘\sec 150^\circsec150∘ (b) csc225∘\csc 225^\circcsc225∘ (c) cot300∘\cot 300^\circcot300∘ Solution (a) sec150∘=1cos150∘=1−32=−23=−233≈−1.1547\sec 150^\circ=\dfrac{1}{\cos 150^\circ}=\dfrac{1}{-\frac{\sqrt3}{2}}=-\dfrac{2}{\sqrt3}=-\dfrac{2\sqrt3}{3}\approx -1{.}1547sec150∘=cos150∘1=−231=−32=−323≈−1.1547. (b) csc225∘=1sin225∘=1−22=−2≈−1.4142\csc 225^\circ=\dfrac{1}{\sin 225^\circ}=\dfrac{1}{-\frac{\sqrt2}{2}}=-\sqrt2\approx -1{.}4142csc225∘=sin225∘1=−221=−2≈−1.4142. (c) cot300∘=cos300∘sin300∘=12−32=−13=−33≈−0.5774\cot 300^\circ=\dfrac{\cos 300^\circ}{\sin 300^\circ}=\dfrac{\frac12}{-\frac{\sqrt3}{2}}=-\dfrac{1}{\sqrt3}=-\dfrac{\sqrt3}{3}\approx -0{.}5774cot300∘=sin300∘cos300∘=−2321=−31=−33≈−0.5774. sec150∘=−233;csc225∘=−2;cot300∘=−33\boxed{\sec 150^\circ=-\tfrac{2\sqrt3}{3};\quad \csc 225^\circ=-\sqrt2;\quad \cot 300^\circ=-\tfrac{\sqrt3}{3}}sec150∘=−323;csc225∘=−2;cot300∘=−33