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

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(B) The expression is given by $E = \prod_{r=1}^{59} \left( 1 - \frac{\cos(60^\circ + r^\circ)}{\cos r^\circ} \right)$.Using the identity $\cos A - \c

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$1$

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(B) The expression is given by $E = \prod_{r=1}^{59} \left( 1 - \frac{\cos(60^\circ + r^\circ)}{\cos r^\circ} ight)$.Using the identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$,we have $1 - \frac{\cos(60^\circ + r^\circ)}{\cos r^\circ} = \frac{\cos r^\circ - \cos(60^\circ + r^\circ)}{\cos r^\circ} = \frac{2 \sin(30^\circ + r^\circ) \sin 30^\circ}{\cos r^\circ} = \frac{\sin(30^\circ + r^\circ)}{\cos r^\circ}$.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,$E = 1$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

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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