Useful Algebra: Using the Conjugate of the Denominator

1. [tex]\((a-3) \times (a+3) = a^2 - 9\)[/tex]
2. [tex]\((3 + t^2) \times (3 - t^2)\)[/tex]
3. [tex]\((\cos \theta + \sin \theta) \times (\cos \theta - \sin \theta)\)[/tex]



Answer :

Let's simplify each of these expressions using the concept of the difference of squares.

The difference of squares formula is:
[tex]\[ (x - y)(x + y) = x^2 - y^2. \][/tex]

### 1. Simplifying [tex]\((a - 3)(a + 3)\)[/tex]

Let's use the difference of squares formula:
[tex]\[ (a - 3)(a + 3) = a^2 - 3^2. \][/tex]
Simplify:
[tex]\[ a^2 - 9. \][/tex]

### 2. Simplifying [tex]\((3 + t^2)(3 - t^2)\)[/tex]

Again, using the difference of squares formula:
[tex]\[ (3 + t^2)(3 - t^2) = (3)^2 - (t^2)^2. \][/tex]
Simplify:
[tex]\[ 9 - t^4. \][/tex]

### 3. Simplifying [tex]\((\cos \theta + \sin \theta)(\cos \theta - \sin \theta)\)[/tex]

Using the difference of squares formula one more time:
[tex]\[ (\cos \theta + \sin \theta)(\cos \theta - \sin \theta) = (\cos \theta)^2 - (\sin \theta)^2. \][/tex]
Notice that [tex]\(\cos^2 \theta - \sin^2 \theta\)[/tex] is the double-angle identity for cosine:
[tex]\[ \cos(2 \theta). \][/tex]

So, the simplified results are:
1. [tex]\(a^2 - 9\)[/tex]
2. [tex]\(9 - t^4\)[/tex]
3. [tex]\(\cos(2\theta)\)[/tex]

Putting it all together:
[tex]\[ \begin{array}{l} (a - 3)(a + 3) \rightarrow a^2 - 9 \\ (3 + t^2)(3 - t^2) \rightarrow 9 - t^4 \\ (\cos \theta + \sin \theta)(\cos \theta - \sin \theta) \rightarrow \cos(2\theta) \end{array} \][/tex]