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(A) $(I)$ The series is $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \dots + \sin^2 90^{\circ}$. There are $18$ terms from $5^{\circ}$ to $85^{\circ}$ plus

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$(I)$-$B$,$(II)$-$D$,$(III)$-$C$,$(IV)$-$A$

Vedclass · Answer

(A) $(I)$ The series is $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \dots + \sin^2 90^{\circ}$. There are $18$ terms from $5^{\circ}$ to $85^{\circ}$ plus $\sin^2 90^{\circ} = 1$. Pairing $\sin^2 	heta + \sin^2(90^{\circ}-	heta) = 1$,we have $8$ pairs plus $\sin^2 45^{\circ} = 0.5$ and $\sin^2 90^{\circ} = 1$. Total $= 8 + 0.5 + 1 = 9.5 = \frac{19}{2}$. Thus,$(I)$-$B$.$(II)$ $	an^2 5^{\circ} \cdot 	an^2 85^{\circ} = 	an^2 5^{\circ} \cdot \cot^2 5^{\circ} = 1$. There are $8$ such pairs and $	an^2 45^{\circ} = 1$. Total $= 1^8 \cdot 1 = 1$. Thus,$(II)$-$D$.$(III)$ $\cos^2 5^{\circ} + \dots + \cos^2 180^{\circ}$. Using $\cos^2 	heta + \cos^2(180^{\circ}-	heta) = 0$ is incorrect; rather $\cos(180^{\circ}-	heta) = -\cos 	heta$,so $\cos^2(180^{\circ}-	heta) = \cos^2 	heta$. The sum is $\sum_{n=1}^{35} \cos^2(5n^{\circ}) + \cos^2 90^{\circ} + \cos^2 180^{\circ}$. This evaluates to $18$. Thus,$(III)$-$C$.$(IV)$ $\cot 	heta + \cot(180^{\circ}-	heta) = \cot 	heta - \cot 	heta = 0$. Pairing terms from $5^{\circ}$ to $175^{\circ}$ gives $0$. $\cot 90^{\circ} = 0$. Total $= 0$. Thus,$(IV)$-$A$.

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

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