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

$\dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\text{cot}^2 A}{\text{1 + cot}^2 A} = 1.$

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

$\Rightarrow \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\dfrac{1}{\text{tan}^2 A}}{1 + \dfrac{1}{\text{tan}^2 A}} \\[1em] = \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\dfrac{1}{\text{tan}^2 A}}{\dfrac{\text{tan}^2 A + 1}{\text{tan}^2 A}} \\[1em] = \dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{1}{\text{tan}^2 A + 1} \\[1em] = \dfrac{\text{tan}^2 A + 1}{\text{tan}^2 A + 1} \\[1em] 1.$

Since, L.H.S. = R.H.S. hence proved that, $\dfrac{\text{tan}^2 A}{\text{1 + tan}^2 A} + \dfrac{\text{cot}^2 A}{\text{1 + cot}^2 A} = 1.$