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Standard Values of Trigonometric Ratios

Question 1 ii...

Question

Question 1 (iii)
Evaluate the following:
(iii) $\frac{c o s 45^{\circ}}{(s e c 30^{\circ} + c o s e c 30^{\circ})}$

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Solution

(iii) $\frac{c o s 45^{\circ}}{(s e c 30^{\circ} + c o s e c 30^{\circ})}$
$= \frac{\frac{1}{\sqrt{2}}}{(\frac{2}{\sqrt{3}} + 2)} = \frac{\frac{1}{\sqrt{2}}}{\frac{(2 + 2 \sqrt{3})}{\sqrt{3}}}$
$= \frac{\sqrt{3}}{\sqrt{2} \times (2 + 2 \sqrt{3})} = \frac{\sqrt{3}}{(2 \sqrt{2} + 2 \sqrt{6})}$
$= \frac{\sqrt{3} \times (2 \sqrt{6} - 2 \sqrt{2})}{(2 \sqrt{6} + 2 \sqrt{2}) \times (2 \sqrt{6} - 2 \sqrt{2})}$
$= \frac{2 \sqrt{3} (\sqrt{6} - \sqrt{2})}{(2 \sqrt{6})^{2} - (2 \sqrt{2})^{2}}$
$= \frac{2 \sqrt{3} (\sqrt{6} - \sqrt{2})}{(24 - 8)} = \frac{2 \sqrt{3} (\sqrt{6} - \sqrt{2})}{16}$
$= \frac{\sqrt{3} (\sqrt{6} - \sqrt{2})}{8} = \frac{(\sqrt{18} - \sqrt{6})}{8} = \frac{(3 \sqrt{2} - \sqrt{6})}{8}$

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Standard Values of Trigonometric Ratios

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