If theta in left(0, frac{pi}{2}right),then le | vedclass

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(C) Let the given determinant be $\Delta$. We know that $(a+b)^2 - (a-b)^2 = 4ab$. Applying the operation $C_1 \rightarrow C_1 - C_2$,the first column

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$0$

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(C) Let the given determinant be $\Delta$. We know that $(a+b)^2 - (a-b)^2 = 4ab$. Applying the operation $C_1 ightarrow C_1 - C_2$,the first column becomes: $C_1(1) = (\sin 	heta + \operatorname{cosec} 	heta)^2 - (\sin 	heta - \operatorname{cosec} 	heta)^2 = 4(\sin 	heta)(\operatorname{cosec} 	heta) = 4(1) = 4$. $C_1(2) = (\cos 	heta + \sec 	heta)^2 - (\cos 	heta - \sec 	heta)^2 = 4(\cos 	heta)(\sec 	heta) = 4(1) = 4$. $C_1(3) = (	an 	heta + \cot 	heta)^2 - (	an 	heta - \cot 	heta)^2 = 4(	an 	heta)(\cot 	heta) = 4(1) = 4$. Now,the determinant is $\Delta = \left|\begin{array}{ccc} 4 & (\sin 	heta-\operatorname{cosec} 	heta)^2 & 2020 \ 4 & (\cos 	heta-\sec 	heta)^2 & 2020 \ 4 & (	an 	heta-\cot 	heta)^2 & 2020 \end{array}ight|$. Since the first column $C_1$ and the third column $C_3$ are proportional (specifically,$C_3 = 505 	imes C_1$),the value of the determinant is $0$.

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