Question

verify the identity.
-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}=-1
to verify the identity, start with
-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}
=-\csc t\square+\cot t\square

Explanation:

Step1: Recall reciprocal and quotient identities

We know that \(\frac{1}{\sin t}=\csc t\) (reciprocal identity) and \(\frac{1}{\tan t}=\cot t\) (since \(\tan t=\frac{\sin t}{\cos t}\), so \(\frac{1}{\tan t}=\frac{\cos t}{\sin t}=\cot t\), quotient identity).

Step2: Simplify each term

For the first term \(-\frac{\csc t}{\sin t}\), since \(\frac{1}{\sin t}=\csc t\), then \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}=-\csc t\times\csc t = -\csc^{2}t\)? Wait, no, wait the problem is to fill in the blanks for \(-\csc t\square+\cot t\square\). Wait, actually, let's re - examine. The original expression is \(-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}\).

We know that \(\frac{1}{\sin t}=\csc t\), so \(\frac{\csc t}{\sin t}=\csc t\times\frac{1}{\sin t}=\csc t\times\csc t=\csc^{2}t\)? No, wait, no. Wait, \(\frac{\csc t}{\sin t}=\csc t\times\frac{1}{\sin t}\), and since \(\csc t = \frac{1}{\sin t}\), then \(\frac{\csc t}{\sin t}=\frac{1}{\sin t}\times\frac{1}{\sin t}=\frac{1}{\sin^{2}t}=\csc^{2}t\). But for the blanks, we need to find the reciprocal of \(\sin t\) and \(\tan t\) in the denominators.

Wait, the first blank: \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}\), and \(\frac{1}{\sin t}=\csc t\)? No, \(\csc t=\frac{1}{\sin t}\), so \(\frac{1}{\sin t}=\csc t\). Wait, no, let's do it step by step.

The first term: \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}\), and \(\frac{1}{\sin t}=\csc t\), so \(-\frac{\csc t}{\sin t}=-\csc t\times\csc t\)? No, that's not right. Wait, maybe the problem is written as \(-\csc t\times\frac{1}{\sin t}+\cot t\times\frac{1}{\tan t}\). Then, since \(\frac{1}{\sin t}=\csc t\) and \(\frac{1}{\tan t}=\cot t\).

So for the first blank (the multiplier of \(-\csc t\)): \(\frac{1}{\sin t}\), and for the second blank (the multiplier of \(\cot t\)): \(\frac{1}{\tan t}\) is wrong. Wait, no, the expression is \(-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}=-\csc t\times\frac{1}{\sin t}+\cot t\times\frac{1}{\tan t}\). And since \(\frac{1}{\sin t}=\csc t\) and \(\frac{1}{\tan t}=\cot t\).

So the first blank (the term multiplied by \(-\csc t\)) is \(\frac{1}{\sin t}\) which is \(\csc t\)? No, wait, let's re - express:

\(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}\), and \(\frac{1}{\sin t}=\csc t\), so \(-\frac{\csc t}{\sin t}=-\csc t\times\csc t\) is incorrect. Wait, maybe the problem has a typo or I misread. Wait, the original problem says "To verify the identity, start with \(-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}=-\csc t\square+\cot t\square\)".

We know that \(\frac{1}{\sin t}=\csc t\), so \(\frac{\csc t}{\sin t}=\csc t\times\frac{1}{\sin t}=\csc t\times\csc t=\csc^{2}t\) is not helpful here. Wait, no, the operation is division. So \(\frac{\csc t}{\sin t}=\csc t\div\sin t\), and \(\sin t=\frac{1}{\csc t}\), so \(\csc t\div\sin t=\csc t\div\frac{1}{\csc t}=\csc t\times\csc t=\csc^{2}t\). But the blanks are after \(-\csc t\) and \(\cot t\).

Wait, maybe the intended approach is:

For the first term: \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}\), and \(\frac{1}{\sin t}=\csc t\), so the first blank is \(\frac{1}{\sin t}\) (but \(\frac{1}{\sin t}=\csc t\)). For the second term: \(\frac{\cot t}{\tan t}=\cot t\times\frac{1}{\tan t}\), and \(\frac{1}{\tan t}=\cot t\), so the second blank is \(\frac{1}{\tan t}\) (but \(\frac{1}{\tan t}=\cot t\)).

Wait, the problem is asking to rewrite \(-\frac{\csc t}{\sin t}+\frac{\cot t}{\tan t}\) as \(-\csc t\square+\cot t\square\). So we can think of \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}\) and \(\frac{\cot t}{\tan t}=\cot t\times\frac{1}{\tan t}\). So the first blank is \(\frac{1}{\sin t}\) (which is \(\csc t\)) and the second blank is \(\frac{1}{\tan t}\) (which is \(\cot t\)).

So:

For the first blank: Since \(\frac{\csc t}{\sin t}=\csc t\times\frac{1}{\sin t}\), so the first blank is \(\frac{1}{\sin t}\) (or \(\csc t\) because \(\frac{1}{\sin t}=\csc t\)).

For the second blank: Since \(\frac{\cot t}{\tan t}=\cot t\times\frac{1}{\tan t}\), and \(\frac{1}{\tan t}=\cot t\), so the second blank is \(\frac{1}{\tan t}\) (or \(\cot t\)).

Answer:

The first blank is \(\frac{1}{\sin t}\) (or \(\csc t\)) and the second blank is \(\frac{1}{\tan t}\) (or \(\cot t\)). If we use the reciprocal identities directly, the first blank is \(\csc t\) (because \(\frac{1}{\sin t}=\csc t\)) and the second blank is \(\cot t\) (because \(\frac{1}{\tan t}=\cot t\)). So the filled - in expression is \(-\csc t\times\csc t+\cot t\times\cot t\)? Wait, no, wait. Wait, \(-\frac{\csc t}{\sin t}=-\csc t\times\frac{1}{\sin t}=-\csc t\times\csc t = -\csc^{2}t\) and \(\frac{\cot t}{\tan t}=\cot t\times\frac{1}{\tan t}=\cot t\times\cot t=\cot^{2}t\). But the problem is to fill in the blanks for \(-\csc t\square+\cot t\square\). So the first blank is \(\csc t\) (because \(\frac{\csc t}{\sin t}=\csc t\times\frac{1}{\sin t}=\csc t\times\csc t\), so \(-\frac{\csc t}{\sin t}=-\csc t\times\csc t\)) and the second blank is \(\cot t\) (because \(\frac{\cot t}{\tan t}=\cot t\times\frac{1}{\tan t}=\cot t\times\cot t\)).

So the first blank: \(\csc t\), the second blank: \(\cot t\)