If A is an acute angle and sin A = $\dfrac{3}{5}$, find all other trigonometric ratios of angle A (using trigonometric identities).

Given sin A = $\dfrac{3}{5}$

sin² A + cos² A = 1

Putting values we get,

$\Rightarrow \Big(\dfrac{3}{5}\Big)^2 + \text{cos}^2A = 1 \\[1em] \Rightarrow \dfrac{9}{25} + \text{cos}^2A = 1 \\[1em] \Rightarrow \text{cos}^2A = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{cos}^2A = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 A = \dfrac{16}{25} \\[1em] \Rightarrow \text{cos} A = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{cos }A = \dfrac{4}{5}.$

sec A = $\dfrac{1}{\text{cos A}} = \dfrac{1}{\dfrac{4}{5}} = \dfrac{5}{4}$.

cosec A = $\dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{3}{5}} = \dfrac{5}{3}.$

1 + tan² A = sec² A

Putting values we get,

$1 + \text{tan}^2A = \Big(\dfrac{5}{4}\Big)^2 \\[1em] 1 + \text{tan}^2A = \dfrac{25}{16} \\[1em] \text{tan}^2A = \dfrac{25}{16} - 1 \\[1em] \text{tan}^2A = \dfrac{25 - 16}{16} \\[1em] \text{tan}^2A = \dfrac{9}{16} \\[1em] \text{tan }A = \sqrt{\dfrac{9}{16}} \\[1em] \text{tan }A = \dfrac{3}{4}.$

1 + cot² A = cosec² A

Putting values we get,

$1 + \text{cot}^2A = \Big(\dfrac{5}{3}\Big)^2 \\[1em] 1 + \text{cot}^2A = \dfrac{25}{9} \\[1em] \text{cot}^2A = \dfrac{25}{9} - 1 \\[1em] \text{cot}^2A = \dfrac{25 - 9}{9} \\[1em] \text{cot}^2A = \dfrac{16}{9} \\[1em] \text{cot }A = \sqrt{\dfrac{16}{9}} \\[1em] \text{cot }A = \dfrac{4}{3}.$

Hence, the value of,

$\text{cos A} = \dfrac{4}{5} \\[1em] \text{tan A} = \dfrac{3}{4} \\[1em] \text{cot A} = \dfrac{4}{3} \\[1em] \text{sec A} = \dfrac{5}{4} \\[1em] \text{cosec A} = \dfrac{5}{3}.$