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Evaluate : $\frac{\sin^2 45^{\circ}}{\text{cosec}^2 30^{\circ} - \tan^2 45^{\circ}}$
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Solution: $\frac{\sin^2 45^{\circ}}{\text{cosec}^2 30^{\circ} - \tan^2 45^{\circ}} = \frac{(\frac{1}{\sqrt{2}})^2}{(2)^2 - (1)^2}$ [1 1/2 marks]
$= \frac{1}{6}$ [1/2 mark]
$= \frac{1}{6}$ [1/2 mark]