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Mathematics Question on Trigonometric Functions

The value of tan⁡330∘sec⁡420∘sin⁡300∘tan⁡135∘sin⁡210∘sec⁡315∘\frac{\tan 330^{\circ} \sec 420^{\circ} \sin 300^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ} } is equal to

A

12\frac{1}{\sqrt{2}}

B

2\sqrt{2}

C

13\frac{1}{\sqrt{3}}

D

3\sqrt{3}

Answer

2\sqrt{2}

Explanation

Solution

tan⁡330∘sec⁡420∘sin⁡300∘tan⁡135∘sin⁡210∘sec⁡315∘\frac{\tan 330^{\circ} \sec 420^{\circ} \sin 300^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ} }
=tan⁡(360−30)∘sec⁡(360+60)∘sin⁡(360−60)∘tan⁡(180−45)∘sin⁡(180+30)∘sec⁡(360−45)∘= \frac{\tan \left(360 -30\right)^{\circ } \sec \left(360+60\right)^{\circ } \sin \left(360 -60\right)^{\circ }}{\tan \left(180 -45\right)^{\circ } \sin\left(180+30\right)^{\circ } \sec\left(360 -45\right)^{\circ } }
=−tan⁡30∘×sec⁡60∘×sin⁡60∘(−tan⁡45∘)(−sin⁡30∘)sec⁡45∘= \frac{- \tan 30^{\circ } \times \sec60^{\circ } \times \sin60^{\circ }}{\left(-\tan 45^{\circ }\right) \left(-\sin 30^{\circ }\right) \sec 45^{\circ } }
=1×2×3×23×2×1×22= \frac{1 \times 2\times \sqrt{3} \times 2}{\sqrt{3} \times 2\times 1\times \sqrt{2}}\sqrt{2}