Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios as defined:

$\dfrac{\text{sin }A}{1 + \text{cot }A} - \dfrac{\text{cos }A}{1 + \text{tan }A}$ = sin A - cos A

The L.H.S of above equation can be written as,

$\Rightarrow \dfrac{\text{sin }A}{(1 + \text{cot }A)} - \dfrac{\text{cos }A}{(1 + \text{tan }A)} \\[1em] \Rightarrow \dfrac{\text{sin }A}{\Big(1 + \dfrac{\text{cos }A}{\text{sin }A}\Big)} - \dfrac{\text{cos }A}{\Big(1 + \dfrac{\text{sin }A}{\text{cos }A}\Big)} \\[1em] \Rightarrow \dfrac{\text{sin }A}{\Big(\dfrac{\text{sin }A + \text{cos }A}{\text{sin }A}\Big)} - \dfrac{\text{cos }A}{\Big(\dfrac{\text{cos }A + \text{sin }A}{\text{cos }A} \Big)} \\[1em] \Rightarrow \dfrac{\text{sin }A \times \text{sin }A}{\text{sin }A + \text{cos }A} - \dfrac{\text{cos }A \times \text{cos }A}{\text{cos }A + \text{sin }A} \\[1em] \Rightarrow \dfrac{\text{sin }^2A - \text{cos }^2A}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \dfrac{(\text{sin }A - \text{cos }A)(\text{sin }A + \text{cos }A)}{\text{sin }A + \text{cos }A}\\[1em] \Rightarrow \text{sin }A - \text{cos }A.$

Since, L.H.S. = R.H.S.

Hence, proved $\dfrac{\text{sin }A}{(1 + \text{cot }A)} - \dfrac{\text{cos }A}{(1 + \text{tan }A)}$ = sin A - cos A