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$\sin 690^{\circ} \times \sec 240^{\circ}=$
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The correct answer is:
$1$
$\sin 690^{\circ} \times \sec 240^{\circ}$
$=\sin \left(1 \times 360^{\circ}+330^{\circ}\right) \times \sec \left(180^{\circ}+60^{\circ}\right)$
$=\sin 330^{\circ} \times\left(-\sec 60^{\circ}\right)$
$=\sin \left(2 \pi-30^{\circ}\right) \times(-2)$
$=-2\left(-\sin 30^{\circ}\right)=(-2)\left(-\frac{1}{2}\right)=1$
$=\sin \left(1 \times 360^{\circ}+330^{\circ}\right) \times \sec \left(180^{\circ}+60^{\circ}\right)$
$=\sin 330^{\circ} \times\left(-\sec 60^{\circ}\right)$
$=\sin \left(2 \pi-30^{\circ}\right) \times(-2)$
$=-2\left(-\sin 30^{\circ}\right)=(-2)\left(-\frac{1}{2}\right)=1$
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