Question

1. using exact values, show that ( 1+cot ^{2} \theta=csc ^{2} \theta ) for ( \theta = 45^{circ} ). 2 marks
2. simplify the following expression using exact values. 3 marks

( cos 60^{circ}(1-sin 30^{circ})+4 \tan 45^{circ}(\tan 60^{circ}-sin 45^{circ}) )

Explanation:

Step1: Calculate left - hand side

When \(\theta = 45^{\circ}\), \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), \(\cot45^{\circ}=\frac{\cos45^{\circ}}{\sin45^{\circ}} = 1\).
Then \(1+\cot^{2}45^{\circ}=1 + 1^{2}=2\).

Step2: Calculate right - hand side

\(\csc\theta=\frac{1}{\sin\theta}\), \(\csc45^{\circ}=\frac{1}{\sin45^{\circ}}=\sqrt{2}\).
So \(\csc^{2}45^{\circ}=(\sqrt{2})^{2}=2\).

Answer:

Since \(1+\cot^{2}45^{\circ}=2\) and \(\csc^{2}45^{\circ}=2\), we have shown that \(1 + \cot^{2}\theta=\csc^{2}\theta\) for \(\theta = 45^{\circ}\).