Question

(sec A – cos A) (cot A + tan A) = sec A tan A

Answer 1

$$\begin{gathered} \left(\sec A-\cos A\right)\left(\cot A+\tan A\right) \\ =\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right) \\ =\left(\frac{1-{\cos }^{2}A}{\cos A}\right)\left(\frac{{\cos }^{2}A+{\sin }^{2}A}{\sin A\cos A}\right) \end{gathered}$$
$$\begin{gathered} =\frac{{\sin }^{2}A}{\cos A}\times \frac{1}{\sin A\cos A}\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{\sin A}{\cos A}\times \frac{1}{\cos A} \\ =\tan A\sec A \end{gathered}$$