If cos A = $\dfrac{9}{41}$; find the value of $\dfrac{1}{\text{sin}^2 A} - \dfrac{1}{\text{tan}^2 A}$.

Given,

cos A = $\dfrac{9}{41}$;

Squaring both sides, we get :

$\Rightarrow \text{cos}^2 A = \Big(\dfrac{9}{41}\Big)^2 \\[1em] \Rightarrow \text{cos}^2 A = \dfrac{81}{1681} \\[1em] \Rightarrow \text{1 - sin}^2 A = \dfrac{81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = 1 - \dfrac{81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = \dfrac{1681 - 81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = \dfrac{1600}{1681} \\[1em] \Rightarrow \text{sin A} = \sqrt{\dfrac{1600}{1681}} \\[1em] \Rightarrow \text{sin A} = \dfrac{40}{41}.$

By formula,

tan A = $\dfrac{\text{sin A}}{\text{cos A}} = \dfrac{\dfrac{40}{41}}{\dfrac{9}{41}} = \dfrac{40}{9}$.

Substituting value of sin A and tan A in $\dfrac{1}{\text{sin}^2 A} - \dfrac{1}{\text{tan}^2 A}$, we get :

$\Rightarrow \dfrac{1}{\text{sin}^2 A} - \dfrac{1}{\text{tan}^2 A} \\[1em] \Rightarrow \dfrac{1}{\Big(\dfrac{40}{41}\Big)^2} - \dfrac{1}{\Big(\dfrac{40}{9}\Big)^2} \\[1em] \Rightarrow \Big(\dfrac{41}{40}\Big)^2 - \Big(\dfrac{9}{40}\Big)^2 \\[1em] \Rightarrow \dfrac{1681}{1600} - \dfrac{81}{1600} \\[1em] \Rightarrow \dfrac{1600}{1600} \\[1em] \Rightarrow 1.$

Hence, $\dfrac{1}{\text{sin}^2 A} - \dfrac{1}{\text{tan}^2 A}$ = 1.