Evaluate:frac{sin 18^{circ}}{cos 72^{circ}} | vedclass

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(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - 	heta) = \cos 	heta$.Given expression: $\frac{\sin 18^{\circ}}{\

Vedclass · Accepted Answer

$1$

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(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - 	heta) = \cos 	heta$.Given expression: $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$.We can write $18^{\circ}$ as $(90^{\circ} - 72^{\circ})$.Therefore,$\sin 18^{\circ} = \sin(90^{\circ} - 72^{\circ}) = \cos 72^{\circ}$.Substituting this into the expression:$\frac{\cos 72^{\circ}}{\cos 72^{\circ}} = 1$.

Evaluate:
$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$

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Evaluate:$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$