Question

1. using exact values, show that ( 1+cot ^{2} \theta=csc ^{2} \theta ) for ( \theta = 45^{circ} ). 2 marks 5. 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

First, find the values of trigonometric functions.
We know that \(\cot45^{\circ}=\frac{\cos45^{\circ}}{\sin45^{\circ}} = 1\), \(\csc45^{\circ}=\frac{1}{\sin45^{\circ}}=\sqrt{2}\).
For the left - hand side \(LHS = 1+\cot^{2}\theta\), when \(\theta = 45^{\circ}\), \(LHS=1 + (\cot45^{\circ})^{2}\).
Substitute \(\cot45^{\circ}=1\) into the formula: \(LHS=1 + 1^{2}=2\).

Step2: Calculate right - hand side

For the right - hand side \(RHS=\csc^{2}\theta\), when \(\theta = 45^{\circ}\), \(RHS = (\csc45^{\circ})^{2}\).
Since \(\csc45^{\circ}=\sqrt{2}\), then \(RHS=(\sqrt{2})^{2}=2\).

Answer:

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