Question
Without using tables, evaluate the following: $\operatorname{cosec}^330^\circ \cos60^\circ \tan^345^\circ \sin^290^\circ \sec^245^\circ \cot30^\circ$.
✓
Answer
$\operatorname{cosec} 30^{\circ} \cos 60^{\circ} \tan ^3 45^{\circ} \sin ^2 90^{\circ} \sec ^2 45^{\circ} \cot 30^{\circ} . $
$ \sin 30^{\circ}=\frac{1}{2} $
$ \operatorname{cosec} 30^{\circ}=2 $
$ \cos 60^{\circ}=\frac{1}{2} $
$ \sec 60^{\circ}=2 $
$ \cos 45^{\circ}=\frac{1}{\sqrt{2}} $
$ \sec 45^{\circ}=\sqrt{2} $
$ \tan 45^{\circ}=1 $
$ \sin 90^{\circ}=1 $
$ \tan 30^{\circ}=\frac{1}{\sqrt{3}} $
$ \Rightarrow \cot 30^{\circ}=\sqrt{3} $
$ \operatorname{cosec} 30^{\circ} \cos 60^{\circ} \tan ^3 45^{\circ} \sin ^2 90^{\circ} \sec ^2 45^{\circ} \cot 30^{\circ} $
$ =(2)^3\left(\frac{1}{2}\right)(1)^3(1)^2(\sqrt{2})^2(\sqrt{3}) $
$ =8 \times \frac{1}{2} \times 2 \times \sqrt{3} $
$ =8 \sqrt{3}$
$ \sin 30^{\circ}=\frac{1}{2} $
$ \operatorname{cosec} 30^{\circ}=2 $
$ \cos 60^{\circ}=\frac{1}{2} $
$ \sec 60^{\circ}=2 $
$ \cos 45^{\circ}=\frac{1}{\sqrt{2}} $
$ \sec 45^{\circ}=\sqrt{2} $
$ \tan 45^{\circ}=1 $
$ \sin 90^{\circ}=1 $
$ \tan 30^{\circ}=\frac{1}{\sqrt{3}} $
$ \Rightarrow \cot 30^{\circ}=\sqrt{3} $
$ \operatorname{cosec} 30^{\circ} \cos 60^{\circ} \tan ^3 45^{\circ} \sin ^2 90^{\circ} \sec ^2 45^{\circ} \cot 30^{\circ} $
$ =(2)^3\left(\frac{1}{2}\right)(1)^3(1)^2(\sqrt{2})^2(\sqrt{3}) $
$ =8 \times \frac{1}{2} \times 2 \times \sqrt{3} $
$ =8 \sqrt{3}$
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