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How do you simplify trigonometric functions like $\tan(x)\sec(x)$ or $\csc(x)\cot(x)$ as well as other equations like $\frac{\tan(x)}{\sec(x)}$ and so on? And could you explain why you are doing the steps so I can understand it a little better and be able to do these on my own. What about $\sec(x)\cot(x)$?

Thank you.

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  • $\begingroup$ What do you mean exactly with simplify? $\tan(x)$ is a definition that you can read on Wikipedia or somewhere else. $\endgroup$ – BIS HD Mar 14 '14 at 13:45
  • $\begingroup$ $$\sec x \cot x = \require{cancel} \dfrac 1{\cancel{\cos x}}\cdot \dfrac{\cancel{\cos x}}{\sin x} = \dfrac 1{\sin x} = \csc x$$ $\endgroup$ – Namaste Mar 14 '14 at 13:56
  • $\begingroup$ @BISHD Its exactly as I have seen the questions on my assignments it is worded "What is a simpler way to write..." or "Which of the following is ... in a simplified form"?" $\endgroup$ – user135435 Mar 14 '14 at 13:58
  • $\begingroup$ Not every trigonometric expression simplifies. $\endgroup$ – steven gregory Mar 4 '16 at 20:17
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It pretty much boils down to using the definitions of $\tan x, \sec x, \csc x, \cot x$.

For example

$$\tan x \sec x = \dfrac{\sin x}{\cos x}\cdot \dfrac 1{\cos x} = \dfrac {\sin x}{\cos^2 x}$$

$$\csc x \cot x = \dfrac{1}{\sin x} \cdot \dfrac{\cos x}{\sin x} = \dfrac{\cos x}{\sin^2 x}$$

$$\sec x \cot x = \require{cancel} \dfrac 1{\cancel{\cos x}}\cdot \dfrac{\cancel{\cos x}}{\sin x} = \dfrac 1{\sin x} = \csc x$$

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  • $\begingroup$ needs more "require{cancel}" :D $\endgroup$ – Guy Mar 14 '14 at 13:37
  • $\begingroup$ @Sabyasachi But nothing cancels in my two examples! ;-) $\endgroup$ – Namaste Mar 14 '14 at 13:38
  • $\begingroup$ OP has a tan(x)/sec(x). do it. I think I saw the require cancel in one of your answers, after that I am using it everywhere. :D $\endgroup$ – Guy Mar 14 '14 at 13:39
  • $\begingroup$ @Sabyasachi Yes, I see that. S/he also asked about $\tan x \sec x$ and $\csc x \cot x$. I saw after first posting my answer that you took on the third function, so didn't want to infringe on "your territory". I think together, we did a smashing job, no? I think we both deserve an upvote (+1);-) $\endgroup$ – Namaste Mar 14 '14 at 13:50
  • $\begingroup$ fair enough. but i think i am infringing on your territory here. you taught me require cancel :D $\endgroup$ – Guy Mar 14 '14 at 13:53
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If you are just working with functions like $\sin$, $\cos$, $\tan$, $\csc$, ... then a good way( at least for a beginner) is to write out all functions in as ratios of $\sin$ and $\cos$. For instance, if you had to simplify $\frac{\tan(x)}{\sec(x)}$, you can write it as $$\large\frac{\frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}}$$

$$\large\require{cancel}{\frac{\frac{\sin(x)}{\cancel{\cos(x)}}}{\frac{1}{\cancel{\cos(x)}}}}$$

$$\sin(x)$$

With time and practice you will get better and learn to do it without breaking it down this explicitly.

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How do you simplify trigonometric functions like $\tan(x)\sec(x)$ or $\csc(x)\cot(x)$ as well as other equations like $\frac{\tan(x)}{\sec(x)}$ and so on?

I suppose it depends on the function at hand and on why you're simplifying it. Simplifying from one perspective can be obfuscating from another, and vice versa. There's rarely a one-size-fits-all approach to this sort of thing. It's more of an art than a science.

Standard identities and "tricks" are always useful, though, like $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y),$$ etc. With enough experience and ingenuity one can sniff out the "right" identity/trick to use and when. [Make sure you understand why these identities hold and then you'll never have to remember them in any great detail.]

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I think that this may be what you are looking for. What you need to do is to remember 4 similar triangles. Note that the pink one can be derived by dividing each side of the blue triangle by $\sin A\cos A$. All that you need to do is to pick the triangle that is most convenient for the problem at hand.

For example, to simplify $\sec A\cot A$ we may want to represent $\cot A$ as $\frac{\text{adjacent side}}{\text{opposite side}}$ in the pink triangle, yeilding $\cot A=\frac{\csc A}{\sec A}$. Then we would simplify the expression as follows. $$\sec A\cot A=\sec A\cdot\frac{\csc A}{\sec A}=\csc A$$ of course we could just as easily used the orange triangle $$\sec A\cot A = \frac{\text{hypotenuse}}{\text{adjacent}}\cdot\cot A=\frac{\csc A}{\cot A}\cdot\cot A=\csc A$$ Finally, lets try to prove that $\tan A+\cot A = \sec A\csc A$. Using the pink triangle as a guide, we can pick off the following relationships. $$\tan A = \frac{\text{opposite}}{\text{adjacent}} =\frac{\sec A}{\csc A}$$ $$\cot A = \frac{\text{adjacent}}{\text{opposite}}= \frac{\csc A}{\sec A}$$ $$\sec^2 A+\csc^2 A=\sec^2 A\csc^2 A\text{ by the Pythagorean theorem}$$ combining these we have $$\tan A+\cot A=\frac{\sec A}{\csc A}+\frac{\csc A}{\sec A}=\frac{\sec^2A+\csc^2A}{\sec A\csc A}=\frac{\sec^2 A\csc^2 A}{\sec A\csc A}=\sec A\csc A$$ I hope this helped :)

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