Answer:
Proof in explanation.
Step-by-step explanation:
[tex]\frac{1}{\sec(x)-1}+\frac{1}{\sec(x)+1}[/tex]
I'm going to find a way to combine the fractions as one.
Multiply the first fraction by [tex]\frac{\sec(x)+1}{\sec(x)+1}[/tex] and multiply the second fraction by [tex]\frac{\sec(x)-1}{\sec(x)-1}[/tex].
[tex]\frac{\sec(x)-1}{(\sec(x)-1)(\sec(x)+1)}+\frac{\sec(x)+1}{(\sec(x)+1)(\sec(x)-1)}[/tex]
[tex]\frac{\sec(x)-1+\sec(x)+1}{(\sec(x)+1)(\sec(x)-1)}[/tex]
The bottom product can easily be determine since when multiplying the conjugate of [tex]a+b[/tex] which [tex]a-b[/tex], we only have to multiply first terms and then last terms giving us [tex]a^2-b^2[/tex].
[tex]\frac{2\sec(x)}{\sec^2(x)-1}[/tex]
Recall the Pythagorean Identity: [tex]1+\tan^2(x)=\sec^2(x)[/tex]
So I can replace [tex]\sec^2(x)-1[/tex] with [tex]\tan^2(x)[/tex]:
[tex]\frac{2\sec(x)}{\tan^2(x)}[/tex]
[tex]2\sec(x) \cdot \frac{1}{\tan^2(x)}[/tex]
[tex]2\sec(x) \cdot \frac{\cos^2(x)}{\sin^2(x)}[/tex]
[tex]2 \sec(x) \cos(x) \cos(x) \frac{1}{\sin^2(x)}[/tex]
[tex]2 (\sec(x) \cos(x)) \cos(x) \frac{1}{\sin^2(x)}[/tex]
[tex]2(1)\cos(x)\csc^2(x)[/tex]
[tex]2\cos(x)\csc^2(x)[/tex]
