Trigonometric Ratios and Identities

Concept:Use trigonometric identities to simplify the numerator and denominator, converting $\csc$ and $\sec$ to $\sin$ and $\cos$, then apply product-to-sum and double-angle formulas.Explanation:Start with $\frac{\sqrt{3}\csc20^\circ-\sec20^\circ}{\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ}$.Since $\csc\theta = 1/\sin\theta$ and $\sec\theta = 1/\cos\theta$, rewrite numerator as $\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ}$.Also $\cos60^\circ=1/2$, so denominator becomes $\cos20^\circ\cos40^\circ\cos80^\circ \cdot \frac12$. Multiply numerator and denominator to simplify.Combine the two fractions in numerator: $\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}$.Thus expression becomes $\frac{2(\sqrt{3}\cos20^\circ-\sin20^\circ)}{\sin20^\circ\cos20^\circ \cdot \cos20^\circ\cos40^\circ\cos80^\circ}$.Now $\sin20^\circ\cos20^\circ = \frac12\sin40^\circ$. Substitute to get denominator $\frac12\sin40^\circ \cdot \cos20^\circ\cos40^\circ\cos80^\circ$.Simplify $\sqrt{3}\cos20^\circ-\sin20^\circ$: write as $2\left(\frac{\sqrt3}{2}\cos20^\circ-\frac12\sin20^\circ\right) = 2(\sin60^\circ\cos20^\circ-\cos60^\circ\sin20^\circ) = 2\sin(60^\circ-20^\circ)=2\sin40^\circ$.Now expression becomes $\frac{2\cdot2\sin40^\circ}{\frac12\sin40^\circ\cos20^\circ\cos40^\circ\cos80^\circ} = \frac{4\sin40^\circ}{\frac12\sin40^\circ\cos20^\circ\cos40^\circ\cos80^\circ}$.Cancel $\sin40^\circ$: $\frac{4}{\frac12\cos20^\circ\cos40^\circ\cos80^\circ} = \frac{8}{\cos20^\circ\cos40^\circ\cos80^\circ}$.Use identity $\cos20^\circ\cos40^\circ\cos80^\circ = \frac18$ (standard result: $\cos20\cos40\cos80 = 1/8$).Thus denominator becomes $\frac18$ and final value is $8 / (1/8) = 64$.Answer:64 (Option B)