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WAEC Mathematics 2017 Paper

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Question : 21

Total: 49

Given that cos 30

^{\circ}

^{\circ}

\;\frac{3}{2}

^{\circ}

^{\circ}

\;\frac{1}{2}

\;\frac{\tan\;60^{\circ}\;-\;1}{1\;-\;\tan\;30^{\circ}}

$\sqrt{3-2}$
2 - $\sqrt{3}$
$\sqrt{3}$
-2

Solution:

Concept:

Substitute the standard trigonometric values and simplify the algebraic fraction.

Explanation:

Given:

\tan 60^{\circ} = \sqrt{3}

and

\tan 30^{\circ} = \frac{1}{\sqrt{3}}

.

Substitute into the expression:

\frac{\tan 60^{\circ} - 1}{1 - \tan 30^{\circ}} = \frac{\sqrt{3} - 1}{1 - \frac{1}{\sqrt{3}}}

Simplify the denominator:

1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}

Therefore:

\frac{\sqrt{3} - 1}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = (\sqrt{3} - 1) \times \frac{\sqrt{3}}{\sqrt{3} - 1} = \sqrt{3}

Answer:

\sqrt{3}

Correct option: C.

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