Simplify sin A / cot A - csc A sin A.

Step-by-step solution

1. 1

   Convert every reciprocal function to sines and cosines. The only reliable first move with $\cot$ and $\csc$ mixed together is to write both in terms of $\sin$ and $\cos$: $\cot A=\frac{\cos A}{\sin A},\qquad \csc A=\frac{1}{\sin A}.$ Everything then lives in one language and the cancellations become visible.

2. 2

   Simplify the first term. Dividing by a fraction means multiplying by its reciprocal: $\frac{\sin A}{\cot A}=\frac{\sin A}{\dfrac{\cos A}{\sin A}}=\sin A\cdot\frac{\sin A}{\cos A}=\frac{\sin^{2}A}{\cos A}.$ Equivalently this is $\sin A\tan A$, which is a useful form to recognise.

3. 3

   Simplify the second term. $\csc A\,\sin A=\frac{1}{\sin A}\cdot\sin A=1,$ valid wherever $\sin A\neq 0$. A reciprocal function multiplied by its own function is always $1$ - that is the whole content of a reciprocal identity.

4. 4

   Combine the two results. $\frac{\sin A}{\cot A}-\csc A\,\sin A=\frac{\sin^{2}A}{\cos A}-1=\frac{\sin^{2}A-\cos A}{\cos A},$ so the expression is $\sin A\tan A-1$ in compact form. No Pythagorean identity helps further, because the numerator mixes $\sin^{2}A$ with a first power of $\cos A$.

5. 5

   State the domain and check a value. The result needs $\sin A\neq 0$ (for $\cot$ and $\csc$ to exist) and $\cos A\neq 0$ (for the quotient), so $A$ must avoid all multiples of $\tfrac{\pi}{2}$. Test at $A=\tfrac{\pi}{3}$: the original is $\dfrac{\sin 60^\circ}{\cot 60^\circ}-\csc 60^\circ\sin 60^\circ=\dfrac{0.8660254}{0.5773503}-1=1.5-1=0.5$, and the simplified form gives $\dfrac{0.75}{0.5}-1=1.5-1=0.5$.