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Been trying this question for over an hour and would like to know how it's done. Thanks.

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\begin{align} & \frac{\frac{\sin\theta}{\cos\theta}}{\cos\theta - \frac{1}{\cos\theta}} \\[10pt] = {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta - 1}{\cos\theta}} \\[10pt] = {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{-(1 - \cos^2\theta)}{\cos\theta}} \\[10pt] = {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{-(\sin^2\theta)}{\cos\theta}} \\[10pt] = {} & \frac{\sin\theta}{\cos\theta} \cdot \frac{\cos\theta}{-\sin^2\theta} \\[10pt] = {} & \frac{-1}{\sin\theta} \\[10pt] = {} & - \csc\theta \end{align}

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

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    Though it is not the question to be answered but still +1 by me – 2017-01-04
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    @THELONEWOLF. Wait what do you mean when you say "thought it is not the question to be answered" kinda confused by that.. also thanks! :) – 2017-01-04
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    I said so because someone downvoted it and I know that it must be someone who thought that it is not a question to be answered, usually highly reputated guys on MSE get irritated from such posts and answers on them. But I m not highly reputated that is why I voted for you. – 2017-01-04
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    Not a post to be answered mean too easy and need no response. – 2017-01-04
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    @THELONEWOLF Ah I see. I like answering these questions since I used to be in the same spot before so I like passing on this knowledge. Thanks for the suggestion anyway - will keep that in mind next time. :) – 2017-01-04
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    I too like answering these questions for the same reason, but when I was typing, I saw your answer flashing and then I did not wanted to put up a duplicate effort. – 2017-01-04
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\begin{align} & \frac{\tan x}{\cos x-\sec x} = \frac{\tan x}{\cos(x)-\frac{1}{\cos(x)}} = \frac{\frac{\sin(x)}{\cos(x)}}{\cos(x)-\frac{1}{\cos(x)}} \\[10pt] = {} &\frac{\sin(x)}{\cos^2(x)-1}=\frac{\sin(x)}{-\sin^2(x)}=-\frac{1}{\sin(x)}=-\csc(x). \end{align}