If sec 4A = operatorname{cosec}(A - 20^{circ} | vedclass

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(B) Given that,$\sec 4A = \operatorname{cosec}(A - 20^{\circ})$.We know that $\sec 	heta = \operatorname{cosec}(90^{\circ} - 	heta)$.Therefore,$\ope

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Given that,$\sec 4A = \operatorname{cosec}(A - 20^{\circ})$.We know that $\sec 	heta = \operatorname{cosec}(90^{\circ} - 	heta)$.Therefore,$\operatorname{cosec}(90^{\circ} - 4A) = \operatorname{cosec}(A - 20^{\circ})$.Equating the angles,we get $90^{\circ} - 4A = A - 20^{\circ}$.Rearranging the terms,$90^{\circ} + 20^{\circ} = A + 4A$.$110^{\circ} = 5A$.$A = \frac{110^{\circ}}{5} = 22^{\circ}$.Thus,the value of $A$ is $22^{\circ}$.

If $\sec 4A = \operatorname{cosec}(A - 20^{\circ})$,where $4A$ is an acute angle,find the value of $A$ (in $^{\circ}$).

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