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Question

(D) Given that $\sec 	heta = \operatorname{cosec} 60^\circ$.We know that $\operatorname{cosec} A = \sec(90^\circ - A)$.Therefore,$\sec 	heta = \sec(

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$\frac{1}{2}$

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(D) Given that $\sec 	heta = \operatorname{cosec} 60^\circ$.We know that $\operatorname{cosec} A = \sec(90^\circ - A)$.Therefore,$\sec 	heta = \sec(90^\circ - 60^\circ) = \sec 30^\circ$.Comparing both sides,we get $	heta = 30^\circ$.Now,substitute $	heta = 30^\circ$ into the expression $2 \cos^2 	heta - 1$:$2 \cos^2 30^\circ - 1 = 2 \left( \frac{\sqrt{3}}{2} ight)^2 - 1$.$= 2 \left( \frac{3}{4} ight) - 1$.$= \frac{3}{2} - 1 = \frac{1}{2}$.

$0 < \theta < 90$ and $\sec \theta = \operatorname{cosec} 60^\circ$,then the value of $2 \cos^2 \theta - 1$ is ........

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