AP DSC School Assistant Math Model Test 1

Concept:Use trigonometric product-to-sum identities to simplify the product of tangents and then substitute known values of special angles.Explanation:Let $E = \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ$. Write each term as $\frac{\sin\theta}{\cos\theta}$:$E = \frac{\sin 6^\circ}{\cos 6^\circ}\cdot\frac{\sin 42^\circ}{\cos 42^\circ}\cdot\frac{\sin 66^\circ}{\cos 66^\circ}\cdot\frac{\sin 78^\circ}{\cos 78^\circ}$.Group the factors in pairs and multiply numerator and denominator by 4:$E = \frac{(2\sin 66^\circ\sin 6^\circ)(2\sin 78^\circ\sin 42^\circ)}{(2\cos 66^\circ\cos 6^\circ)(2\cos 78^\circ\cos 42^\circ)}$.Apply $2\sin A\sin B = \cos(A-B)-\cos(A+B)$ and $2\cos A\cos B = \cos(A+B)+\cos(A-B)$:$E = \frac{[\cos(60^\circ)-\cos(72^\circ)][\cos(36^\circ)-\cos(120^\circ)]}{[\cos(60^\circ)+\cos(72^\circ)][\cos(36^\circ)+\cos(120^\circ)]}$.Now use $\cos 72^\circ = \sin 18^\circ = \frac{\sqrt5-1}{4}$, $\cos 36^\circ = \frac{\sqrt5+1}{4}$, $\cos 60^\circ = \frac12$, and $\cos 120^\circ = -\frac12$. Substitute:$E = \frac{\left(\frac12 - \frac{\sqrt5-1}{4}\right)\left(\frac{\sqrt5+1}{4} + \frac12\right)}{\left(\frac12 + \frac{\sqrt5-1}{4}\right)\left(\frac{\sqrt5+1}{4} - \frac12\right)}$.Simplify each bracket: $\frac12 = \frac24$. Numerator: first factor = $\frac{2 - (\sqrt5-1)}{4} = \frac{3-\sqrt5}{4}$; second factor = $\frac{\sqrt5+1+2}{4} = \frac{3+\sqrt5}{4}$. Denominator: first factor = $\frac{2+\sqrt5-1}{4} = \frac{1+\sqrt5}{4}$; second factor = $\frac{\sqrt5+1-2}{4} = \frac{\sqrt5-1}{4}$.Thus $E = \frac{(3-\sqrt5)(3+\sqrt5)}{(1+\sqrt5)(\sqrt5-1)} = \frac{9-5}{5-1} = \frac{4}{4} = 1$.Answer:The value is $1$, which corresponds to option B.