Question

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

Explanation:

Step1: Calculate the value of \(1+\cot^{2}\theta\)

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

Step2: Calculate the value of \(\csc^{2}\theta\)

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

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