(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\) (b) Given that \(\sin x = \frac{2}{3}\), evaluate, lea...
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(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
Answer Details
(a) Convert to improper fractions.
Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).
Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).
\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]
(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so
\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]
\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]
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