If A is an acute angle and sec A = $\dfrac{17}{8}$, find all other trigonometric ratios of angle A (using trigonometric identities).

Given sec A = $\dfrac{17}{8}$

⇒ cos A = $\dfrac{1}{\text{sec A}} = \dfrac{1}{\dfrac{17}{8}} = \dfrac{8}{17}$.

sin² A + cos² A = 1

Putting values we get,

$\Rightarrow \text{sin}^2A + \Big(\dfrac{8}{17}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2A + \dfrac{64}{289} = 1 \\[1em] \Rightarrow \text{sin}^2A = 1 - \dfrac{64}{289} \\[1em] \Rightarrow \text{sin}^2A = \dfrac{289 - 64}{289} \\[1em] \Rightarrow \text{sin}^2 A = \dfrac{225}{289} \\[1em] \Rightarrow \text{sin } A = \sqrt{\dfrac{225}{289}} \\[1em] \Rightarrow \text{sin }A = \dfrac{15}{17}.$

cosec A = $\dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{15}{17}} = \dfrac{17}{15}.$

1 + tan² A = sec² A

Putting values we get,

$1 + \text{tan}^2A = \Big(\dfrac{17}{8}\Big)^2 \\[1em] 1 + \text{tan}^2A = \dfrac{289}{64} \\[1em] \text{tan}^2A = \dfrac{289}{64} - 1 \\[1em] \text{tan}^2A = \dfrac{289 - 64}{64} \\[1em] \text{tan}^2A = \dfrac{225}{64} \\[1em] \text{tan }A = \sqrt{\dfrac{225}{64}} \\[1em] \text{tan }A = \dfrac{15}{8}.$

1 + cot² A = cosec² A

Putting values we get,

$1 + \text{cot}^2A = \Big(\dfrac{17}{15}\Big)^2 \\[1em] 1 + \text{cot}^2A = \dfrac{289}{225} \\[1em] \text{cot}^2A = \dfrac{289}{225} - 1 \\[1em] \text{cot}^2A = \dfrac{289 - 225}{225} \\[1em] \text{cot}^2A = \dfrac{64}{225} \\[1em] \text{cot }A = \sqrt{\dfrac{64}{225}} \\[1em] \text{cot } A = \dfrac{8}{15}.$

Hence, the value of,

$\text{sin A} = \dfrac{15}{17} \\[1em] \text{cos A} = \dfrac{8}{17} \\[1em] \text{tan A} = \dfrac{15}{8} \\[1em] \text{cot A} = \dfrac{8}{15} \\[1em] \text{cosec A} = \dfrac{17}{15}.$