Showing $1/\left(\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)}\right)$

Question

I'm having a hard time with some basic rule of fractions that I just don't get. I'm told to show that $\frac{1}{\sec(x)+\tan(x)} = \sec(x)-\tan(x)$. I start with the left hand side and convert it to

$$\frac{1}{\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)}}.\tag{Step 1}\label1$$

From here I would normally remove the main 1 in the numerator and invert the fractional denominators by performing the division.

So

$$\frac{1}{1} \times \frac{\cos(x)}{1} + \frac{\cos(x)}{\sin(x)}\tag{Step 2}.\label2$$

Doing this lead me down a path that just doesn't work. The book answer shows them taking the denominator of each fraction in the denominator in \ref{1} (meaning the $\cos(x)$) and just multiply it out of the main fraction.

So they show as: $$\frac{\cos(x)}{1 + \sin(x)} \tag{Step 3}.$$

But this is driving me crazy because their \ref{2} can not be an equality for my step to. I've tried it with basic fractions and just can't see it.
Can someone tell me which one is right?

Answer 1

Note that $$\frac{1}{\frac{1}{a}+\frac{b}{a}} \neq \frac{a}{1}+\frac{a}{b}. \tag{1} $$ To see that, put $a=b=1$ for example. Rather, by using the fact that $$\frac{1}{a}+\frac{b}{a} = \frac{1+b}{a}, \tag{2}$$ we can write $(1)$ as $$\frac{1}{\frac{1}{a}+\frac{b}{a}} = \frac{1}{\frac{1+b}{a}} =\frac{a}{1+b}. \tag{3}$$ In your case, $a = \cos(x)$ and $b=\sin(x)$.

Answer 2

$\dfrac{1}{\sec(x)+\tan(x)} = \dfrac{\sec(x)-\tan(x)}{\sec^2(x)-\tan^2(x)} = \dfrac{\sec(x)-\tan(x)}{1}=\sec(x)-\tan(x)$

If you need to justify the denominator:

$(\sec(x)-\tan(x))(\sec(x)+\tan(x)) = \sec^2(x)-\tan^2(x) = \dfrac{1-\sin^2(x) }{\cos^2(x)} = \dfrac{\cos^2(x) }{\cos^2(x)} =1$

Answer 3

$$\frac{1}{\sec x+ \tan x} =\frac{1}{\frac1{\cos x}+ \frac{\sin x}{\cos x}} =\frac{\cos x}{1+ \sin x}=\frac{\cos x}{1+ \sin x} \cdot \frac{1- \sin x}{1- \sin x} =\\=\frac{\cos x(1-\sin x)}{1- \sin^2 x} =\frac{\cos x-\sin x\cos x}{\cos^2 x} = \sec x -\tan x $$

Answer 4

$$\frac1{\sec x-\tan x}=\sec x+\tan x$$ can be written $$\sec^2x-\tan^2x=1$$

or

$$\frac{1-\sin^2x}{\cos^2x}=1.$$