Question

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

Explanation:

Step1: Calculate the left - hand side

We know that \(\cot\theta=\frac{\cos\theta}{\sin\theta}\). For \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\) and \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\). Then \(\cot45^{\circ}=\frac{\cos45^{\circ}}{\sin45^{\circ}} = 1\).
So \(1+\cot^{2}45^{\circ}=1 + 1^{2}=2\).

Step2: Calculate the right - hand side

We know that \(\csc\theta=\frac{1}{\sin\theta}\). For \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\csc45^{\circ}=\frac{1}{\sin45^{\circ}}=\sqrt{2}\).
And \(\csc^{2}45^{\circ}=(\sqrt{2})^{2}=2\).

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 \(\theta = 45^{\circ}\).