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.

  • $\begingroup$ How did you get 2 on the top? $\endgroup$ – tabstop Mar 3 '14 at 23:53
  • $\begingroup$ So, specifically, you want to prove that your first expression is equal to $2\cot(x)\csc(x)$? $\endgroup$ – Jack M Mar 3 '14 at 23:54

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

  • $\begingroup$ How were you able to change $2sec(x)$ to $2cos(x)$ and $tan^2(x)$ to $sin^2(x)$? $\endgroup$ – Tanner Mar 4 '14 at 0:36
  • $\begingroup$ $$\dfrac{2\sec(x)}{\tan^2(x)}=\dfrac{2}{\tan^2(x)\cos(x)}=\dfrac{2\cos^2(x)}{\sin^2(x) \cos(x)}=\dfrac{2\cos(x)}{\sin^2(x)}=\dfrac{2\cot(x)}{\sin(x)}$$The definition of $\sec(x)$ as $\frac{1}{\cos(x)}$ and the fact that $\frac{1}{\frac{b}{a}}=\frac{a}{b}$ was used. $\endgroup$ – user122283 Mar 4 '14 at 0:44
  • $\begingroup$ Thanks for explaining that to me it; it really helped! $\endgroup$ – Tanner Mar 4 '14 at 0:50
  • $\begingroup$ @Tanner Anything for a fellow mathematician! $\endgroup$ – user122283 Mar 4 '14 at 0:50

The numerator should be $2\sec x$.

$$\require{cancel}$$ $$\begin{align}\dfrac{1}{\sec(x)+1} + \dfrac{1}{\sec(x)-1} & = \dfrac{\sec x - 1 + \sec x + 1}{\sec^2 x - 1} \\ \\ &= \dfrac{2\sec x}{\underbrace{\sec^2 x - 1}_{\tan^2 x}} \\ \\ &= \dfrac{\frac 2{\cos x}}{\frac {\sin^2 x}{\cos^2 x}}\\ \\ & = \dfrac 2{\frac{\sin^2 x}{\cos x}} \\ \\ &= \dfrac {2\cos x}{(\sin x)(\sin x)} \\ \\ &= 2\cdot \dfrac{\cos x }{\sin x}\cdot \frac 1{\sin x}\\ \\ & = 2\cot x\cos x\end{align}$$


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.


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