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I am having trouble solving a trigonometry identity problem. Specifically, I need to solve:

$\dfrac{\sec(x)\sin(x)}{\tan(x)\cot(x)} = \sin^2x$

I tried solving this by turning $\sin^2x$ into it's half-angle equivalent, starting from the original trigonometric identity to get to my answer backwards, etc etc etc for the last 2 hours. Any suggestions would be appreciated.

Edit: So a few of you asked if my problem had some sort of typing error. This isn't my problem specifically; this is a friend of mine's math problem, so I currently have no way of verifying if this was correctly written to me and will check with him tomorrow. After double-checking with the image he sent me, this is correctly written to the information he gave me.

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    $\begingroup$ $\tan x\cot x=?$ And it looks like you can cancel a $\sin x$ term from both sides as well. But I think there might be a copy error somewhere... $\endgroup$
    – abiessu
    Apr 2, 2015 at 3:22
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    $\begingroup$ As written, you are asking to solve $1 = \sin x \cos x$. Is this what you indented? $\endgroup$
    – davidlowryduda ♦
    Apr 2, 2015 at 3:23
  • $\begingroup$ So for maximum clarity, this is not a homework problem of my own - I am a student of applied mathematics at college and this problem was brought to me from a friend of mine in Precalculus. So, to abiessu; I have no direct way of verifying if there was a copy error for the time being, and to mixedmath; this is what I intended, yes. $\endgroup$
    – Joe D
    Apr 2, 2015 at 3:28
  • $\begingroup$ I think abiessu's right. Are you sure you copied the formula correctly?I worked it out and got tan x. So I'd check that out again if I were you. $\endgroup$
    – Mathemagician1234
    Apr 2, 2015 at 3:59
  • $\begingroup$ @JoeD This isn't exactly an identity. An identity is true $\forall x$. What you're asking is to solve a trigonometric equation. $\endgroup$
    – Arpan
    Apr 2, 2015 at 4:16

1 Answer 1

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First note that $$\frac{\sec(x)\sin(x)}{\tan(x)\cot(x)}= \frac{\frac{1}{\cos(x)}\sin(x)}{\tan(x)\frac{1}{\tan(x)}} =\tan(x)$$ This is clearly not an identity because in general $$ \tan(x)\not =\sin^2(x) $$

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