Let E = left( {1 - frac{{cos 61^circ}}{{cos 1 | vedclass

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(B) Given $E = \prod_{r=1}^{59} \left( {1 - \frac{{\cos(60^\circ + r^\circ)}}{{\cos r^\circ}}} \right)$.Simplifying the term inside the product:$1 - \

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$1$

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(B) Given $E = \prod_{r=1}^{59} \left( {1 - \frac{{\cos(60^\circ + r^\circ)}}{{\cos r^\circ}}} ight)$.Simplifying the term inside the product:$1 - \frac{{\cos(60^\circ + r^\circ)}}{{\cos r^\circ}} = \frac{{\cos r^\circ - \cos(60^\circ + r^\circ)}}{{\cos r^\circ}}$.Using the formula $\cos A - \cos B = -2 \sin \frac{{A+B}}{2} \sin \frac{{A-B}}{2}$:$\cos r^\circ - \cos(60^\circ + r^\circ) = 2 \sin \frac{{r^\circ + 60^\circ + r^\circ}}{2} \sin \frac{{60^\circ + r^\circ - r^\circ}}{2} = 2 \sin(30^\circ + r^\circ) \sin 30^\circ$.Since $\sin 30^\circ = \frac{1}{2}$,the expression becomes $2 \cdot \frac{1}{2} \sin(30^\circ + r^\circ) = \sin(30^\circ + r^\circ)$.Thus,$E = \prod_{r=1}^{59} \frac{{\sin(30^\circ + r^\circ)}}{{\cos r^\circ}} = \frac{{\sin 31^\circ \cdot \sin 32^\circ \dots \sin 89^\circ}}{{\cos 1^\circ \cdot \cos 2^\circ \dots \cos 59^\circ}}$.Since $\sin(90^\circ - 	heta) = \cos 	heta$,the numerator is $\cos 59^\circ \cdot \cos 58^\circ \dots \cos 1^\circ$,which cancels exactly with the denominator.Therefore,$E = 1$.

Let $E = \left( {1 - \frac{{\cos 61^\circ}}{{\cos 1^\circ}}} \right) \left( {1 - \frac{{\cos 62^\circ}}{{\cos 2^\circ}}} \right) \dots \left( {1 - \frac{{\cos 119^\circ}}{{\cos 59^\circ}}} \right)$,then $E$ is equal to:

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