If cos X = $\dfrac{2}{3}$ then tan X is equal to:
By trigonometry identities, we know:
$1 + tan^{2}X $= $sec^{2}X$
And sec X = $\dfrac{1}{cos X}$ =$ \dfrac{1}{(\dfrac{2}{3})}$ =$ \dfrac{3}{2}$
Hence,
$1 + tan^{2}X$ = $(\dfrac{3}{2})^{2}$ =$ \dfrac{9}{4}$
$tan^{2}X$ = $(\dfrac{9}{4}) $– 1 = $\dfrac{5}{4}$
tan X = $\dfrac{\sqrt{5}}{2}$
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