Prove the following:

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

Solving L.H.S. of the equation : $\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A}$ = 1.

$\phantom{=} \dfrac{1}{1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}} + \dfrac{1}{1 + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \dfrac{1}{\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}} + \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \dfrac{\text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} + \dfrac{\text{sin}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = 1.$

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

Hence, proved that $\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A}$ = 1.