Prove the following identities :

tan A - cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Solving L.H.S. of the equation :

$\Rightarrow \text{tan A - cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{cos}^2 A}{\text{cos A sin A}}$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{1 - \text{cos}^2 A - \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{2 cos}^2 A}{\text{sin A cos A}}.$

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

Hence, proved that tan A - cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$