Prove that: (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A.

Hint: We will solve the LHS of the expression first. We will multiply (1 + cot A + tan A) with (sin A – cos A). We will require the following formulas to solve this question, $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$.

Complete step by step solution:
It is given in the question that we have to prove the expression, (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A. We will start by solving the LHS first. So, we can write the LHS as,
(1 + cot A + tan A) (sin A – cos A)
On expanding, we get,
(sin A – cos A) + cot A (sin A – cos A) + tan A (sin A – cos A)
We can write it as,
sin A – cos A + cot A sin A – cot A cos A + tan A sin A – tan A cos A
Now, we know that we can write cot A as $\dfrac{\cos A}{\sin A}$ and tan A as $\dfrac{\sin A}{\cos A}$. So, we get,
$\begin{align}
  & \sin A-\cos A+\dfrac{\cos A}{\sin A}\sin A-\cot A\cos A+\tan A\sin A-\dfrac{\sin A}{\cos A}\cos A \\
 & \sin A-\cos A+\cos A-\cot A\cos A+\tan A\sin A-\sin A \\
 & \sin A-\sin A+\cos A-\cos A-\cot A\cos A+\tan A\sin A \\
 & \tan A\sin A-\cot A\cos A \\
\end{align}$
Therefore, we have obtained the RHS and hence proved that (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A.

Note: The possible mistake that one can make in this question is while expanding the LHS, that is (1 + cot A + tan A) (sin A – cos A). Some students may forget to take the negative sign and this will lead to errors. One can also solve this question starting from the RHS, but it will be a little complicated, so it is better to start from the LHS in this question.