How do you express $\dfrac{\sin A\sec A\cot A}{\tan A}$ in terms of sine and cosine?
I have simplified using $\sec(A)$ as $\cos^{-1}(A)$ and also $\cot(A)$ as $\dfrac{\cos(A)}{\sin(A)}$, and appear to end up with $\dfrac{\sin(A)}{\cos(A)}$, which is simply $\tan(A)$.
On
$$ \frac{\sin A\sec A\cot A}{\tan A} = \frac{\sin A\cdot\dfrac 1 {\cos A}\cdot \dfrac {\cos A} {\sin A}}{\dfrac{\sin A}{\cos A}} $$
That expresses it in terms of sine and cosine. After that, you can do some simplifying. Dividing by $\sin A/\cos A$ is the same as multiplying by $\cos A/\sin A$, so you get $$ \sin A\cdot\dfrac 1 {\cos A}\cdot \dfrac {\cos A} {\sin A} \cdot \frac{\cos A}{\sin A}. $$ Then, doing a couple of cancelations of factors common to the numerator and the denominator, you get $$ \frac{\cos A}{\sin A}. $$ That expresses it in terms of the sine and the cosine of $A$.
You can also write it as $\cot A$.
$$\sin(A)\frac{\sec(A)}{\tan{A}}\cot{A}=\frac{\sin(A)}{\cos(A)}\frac{1}{\tan(A)}\cot(A)=\cot(A)=\frac{\cos(A)}{\sin(A)}$$