Precalc Trig Identities I started off with this problem:
$$ \dfrac{1}{\sec(x)+1} + \dfrac{1}{\sec(x)-1}$$
I attempted to solve it by multiplying the denominators, which gave me this:
$$ \dfrac{2}{\sec^2(x)-1} $$
I then came to the answer of $2\tan(x)$, however I know the answer is actually $2\cot(x)\csc(x)$.
Can someone please point me in the right direction? I'm assuming I made a silly mistake somewhere.
 A: The numerator should be $2\sec x$.
$$\require{cancel}$$
$$\begin{align}\dfrac{1}{\sec(x)+1} + \dfrac{1}{\sec(x)-1} & = \dfrac{\sec x - 1 + \sec x + 1}{\sec^2 x - 1} \\ \\ 
&= \dfrac{2\sec x}{\underbrace{\sec^2 x - 1}_{\tan^2 x}} \\ \\ 
&= \dfrac{\frac 2{\cos x}}{\frac {\sin^2 x}{\cos^2 x}}\\ \\
& = \dfrac 2{\frac{\sin^2 x}{\cos x}} \\ \\ &= \dfrac {2\cos x}{(\sin x)(\sin x)} \\ \\ 
&= 2\cdot \dfrac{\cos x }{\sin x}\cdot \frac 1{\sin x}\\ \\ & = 2\cot x\cos x\end{align}$$
A: $$ \dfrac{1}{\sec(x)+1} + \dfrac{1}{\sec(x)-1}\\
\dfrac{\sec(x)+1+\sec(x)-1}{\sec^2(x)-1}=\dfrac{2\sec(x)}{\sec^2(x)-1}=\dfrac{2\sec(x)}{\tan^2(x)}=\dfrac{2\cos(x)}{\sin^2(x)}=\dfrac{2\cot(x)}{\sin(x)}=\\
2\cot(x)\csc(x)$$
A: Getting rid of the various abbreviations, we want to show:
$$\frac{1}{1/\cos x + 1}+\frac{1}{1/\cos x - 1}=\frac{2}{\tan x\sin x}$$
And plugging in the definition of $\tan$:
$$\frac{1}{1/\cos x + 1}+\frac{1}{1/\cos x - 1}=\frac{2\cos x}{\sin x\sin x}$$
From here it's just a matter of simplifying the left hand side, and a single application of the Pythagorean theorem.
