Match the items of List - I with those of the entires of List - II $List - I$ $\begin{aligned} & \text { (I)…
Match the items of List - I with those of the entires of List - II
$List - I$
$\begin{aligned} & \text { (I) } \sin ^2 5^{\circ}+\sin ^2 10^{\circ}+ \\ & \sin ^2 15^{\circ}+\ldots+\sin ^2 90^{\circ}=\end{aligned}$
$\begin{aligned} & \text { (II) } \tan ^2 5^{\circ} \cdot \tan ^2 10^{\circ} \text {. } \\ & \tan ^2 15^{\circ} \ldots \tan ^2 85^{\circ}= \\ & \end{aligned}$
$\begin{aligned} & \text { (III) } \cos ^2 5^{\circ}+\cos ^2 10^{\circ} \\ & +\cos ^2 15^{\circ}+\ldots+\cos ^2 180^{\circ}=\end{aligned}$
$\begin{gathered}\text { (IV) } \cot 5^{\circ}+\cot 10^{\circ}+\cot 15^{\circ} \\ +\ldots .+\cot 175^{\circ}=\end{gathered}$
$List - II$
(A) $0$
(B) $\frac{19}{2}$
(C) $18$
(D) $1$
(E) $-1$
- (I) - (B), (II) - (D), (III) - (C), (IV) - (A)
- (I) - (B), (II) - (E), (III) - (A), (IV) - (C)
- (I) - (B), (II) - (C), (III) - (A), (IV) - (D)
- (I) - (C), (II) - (B), (III) - (D), (IV) - (E)
Solution
$
\begin{aligned}
& \text { } \begin{array}{l}
\text { (I) } \sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\sin ^2 15^{\circ}+\ldots+\sin ^2 90^{\circ} \\
=\sin ^2 90^{\circ}+\left(\sin ^2 5^{\circ}+\sin ^2 85^{\circ}\right)+\left(\sin ^2 10^{\circ}+\sin ^2 80^{\circ}\right)+ \\
\left(\sin ^2 15^{\circ} \sin ^2 75^{\circ}\right)+\ldots+\left(\sin ^2 40^{\circ}+\sin ^2 50^{\circ}\right)+\sin ^2 45^{\circ}
\end{array} \\
& =(1)^2+\left(\sin ^2 5^{\circ}+\cos ^2 5^{\circ}\right)+\left(\sin ^2 10^{\circ}+\cos ^2 10^{\circ}\right) \\
& +\left(\sin ^2 15^{\circ}+\cos ^2 15^{\circ}\right)+\ldots+\left(\sin ^2 40^{\circ}+\cos ^2 40^{\circ}\right)+\left(\frac{1}{\sqrt{2}}\right)^2 \\
& \therefore \sin ^2\left(90^{\circ}-5^{\circ}\right)=\cos ^2 5^{\circ} \text { and so on }
\end{aligned}
$
$
\begin{gathered}
\text { (II) } \tan ^2 5^{\circ} \tan ^2 10^{\circ} \cdot \tan ^2 15^{\circ} \ldots \tan ^2 85^{\circ} \\
=\left[\left(\tan ^2 5^{\circ} \cdot \tan ^2 85^{\circ}\right)\left(\tan ^2 10^{\circ} \cdot \tan ^2 80\right) \ldots\right. \\
\left.\left(\tan ^2 40^{\circ} \cdot \tan ^2 50^{\circ}\right)\right] \cdot \tan ^2 45^{\circ} \\
=\left[\left(\tan ^2 5^{\circ} \cdot \cot ^2 5^{\circ}\right) \cdot\left(\tan ^2 10^{\circ} \cdot \cot ^2 10^{\circ}\right) \ldots\right. \\
=[1.1 \ldots 1] \cdot 1=1
\end{gathered}
$
Hence (II) $\rightarrow$ (D)
$
\begin{aligned}
& \text { (III) } \cos ^2 5^{\circ}+\cos ^2 10^{\circ}+\cos ^2 15^{\circ}+\ldots+\cos ^2 180^{\circ} \\
& \Rightarrow\left(\cos ^2 5^{\circ}+\cos ^2 95^{\circ}\right)+\left(\cos ^2 10^{\circ}+\cos ^2 100^{\circ}\right)+\ldots \\
& +\left(\cos ^2 90^{\circ}+\cos ^2 180^{\circ}\right) \\
& \Rightarrow\left(\cos ^2 5^{\circ}+\sin ^2 5^{\circ}\right)+\left(\cos ^2 10^{\circ}+\sin ^2 10^{\circ}\right)+\ldots \\
& +\left(\cos ^2 90^{\circ}+\sin ^2 90^{\circ}\right) \\
& \Rightarrow \overbrace{1+1+1+\ldots 18 \text { terms }}^{20 \text { terms }} \\
& \Rightarrow 18 \\
&
\end{aligned}
$
$
\begin{aligned}
& \text { Hence (III) } \rightarrow \text { (C) } \\
& \begin{array}{l}
\text { (IV) } \cot 5^{\circ}+\cot 10^{\circ}+\cot 15^{\circ}+\ldots+\cot 175^{\circ} \\
=\left(\cot 5^{\circ}+\cot 175^{\circ}\right)+\left(\cot 10^{\circ}+\cot 170^{\circ}\right)+\ldots \\
\quad+\left(\cot 85^{\circ}+\cot 95^{\circ}\right)+\cot 90^{\circ} \\
=\left(\cot 5^{\circ}-\cot 5^{\circ}\right)+\left(\cot 10^{\circ}-\cot 10^{\circ}\right)+\ldots+ \\
=0+0+\ldots+0=0 \quad\left(\cot 85^{\circ}-\cot 85^{\circ}\right)+\cot 90^{\circ}
\end{array}
\end{aligned}
$
Hence (IV) $\rightarrow$ (A)
Asked in: AP EAMCET 2023 (19 May Shift 1)
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