If 3 cot A = 4, check whether $\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A}$ = cos² A - sin² A or not.

Given,

⇒ 3 cot A = 4

⇒ cot A = $\dfrac{4}{3}$

Let us draw a right angle triangle ABC.

We know that,

cot A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}}$

Substituting values, we get :

$\dfrac{\text{AB}}{\text{BC}} = \dfrac{4}{3}$

Let AB = 4k and BC = 3k.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (4k)² + (3k)²

⇒ AC² = 16k² + 9k²

⇒ AC² = 25k²

⇒ AC = $\sqrt{25k^2}$ = 5k.

We know that,

tan A = $\dfrac{1}{\text{cot A}} = \dfrac{1}{\dfrac{4}{3}} = \dfrac{3}{4}$.

Substituting value of tan A in $\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A}$

$\Rightarrow \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} \\[1em] \Rightarrow \dfrac{1 - \Big(\dfrac{3}{4}\Big)^2}{1 + \Big(\dfrac{3}{4}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{9}{16}}{1 + \dfrac{9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{16 - 9}{16}}{\dfrac{16 + 9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}} \\[1em] \Rightarrow \dfrac{7}{25}.$

We know that,

cos A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4k}{5k} = \dfrac{4}{5}$.

sin A = $\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3k}{5k} = \dfrac{3}{5}$.

Substituting value of cos A and sin A in cos² A - sin² A, we get :

$\Rightarrow \Big(\dfrac{4}{5}\Big)^2 - \Big(\dfrac{3}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{16}{25} - \dfrac{9}{25} \\[1em] \Rightarrow \dfrac{7}{25}.$

Hence, proved that $\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A}$ = cos² A - sin² A.