2A is the measure of an acute angle and sec 2 | vedclass

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Question

(A) Given that $\sec 2A = \operatorname{cosec}(A - 42^\circ)$.We know that $\sec 	heta = \operatorname{cosec}(90^\circ - 	heta)$.Therefore,$\sec 2A

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) Given that $\sec 2A = \operatorname{cosec}(A - 42^\circ)$.We know that $\sec 	heta = \operatorname{cosec}(90^\circ - 	heta)$.Therefore,$\sec 2A = \operatorname{cosec}(90^\circ - 2A)$.Substituting this into the given equation:$\operatorname{cosec}(90^\circ - 2A) = \operatorname{cosec}(A - 42^\circ)$.Equating the angles:$90^\circ - 2A = A - 42^\circ$.Rearranging the terms to solve for $A$:$90^\circ + 42^\circ = A + 2A$.$132^\circ = 3A$.$A = \frac{132^\circ}{3} = 44^\circ$.

$2A$ is the measure of an acute angle and $\sec 2A = \operatorname{cosec}(A - 42^\circ)$,then the value of $A$ is $\ldots \ldots \ldots \ldots$ (in $^\circ$)

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