PLEASE HELP ME WITH THIS!!! will mark brainliest

[tex] \sin^2 \theta + \cos^2 \theta = 1 [/tex]

[tex] \dfrac{\sin^2 \theta}{\cos^2 \theta} + \dfrac{\cos^2 \theta}{\cps^2 \theta} = \dfrac{1}{\cos^2 \theta} [/tex]

[tex] \tan^2 \theta + 1 = \sec^2 \theta [/tex]

Use the identity in the form:

[tex] \tan^2 A + 1 = \sec^2 A [/tex]

You are given

[tex] \sec A = \dfrac{4}{3} [/tex]

Plug it in and solve for the tangent square.

[tex] \tan^2 A + 1 = \sec^2 A [/tex]

[tex] \tan^2 A + 1 = \left( \dfrac{4}{3} \right)^2 [/tex]

[tex] \tan^2 A + 1 = \dfrac{16}{9} [/tex]

[tex] \tan^2 A = \dfrac{16}{9} - 1 [/tex]

[tex] \tan^2 A = \dfrac{16}{9} - \dfrac{9}{9} [/tex]

[tex] \tan^2 A = \dfrac{7}{9} [/tex]

Since cot A = 1/(tan A),

[tex] \cot^2 A = \dfrac{9}{7} [/tex]

[tex] \cot A = \dfrac{3}{\sqrt{7}} [/tex]

[tex] \cot A = \dfrac{3\sqrt{7}}{7} [/tex]