Solution
$$\frac{Tan x}{Cot x}=Tan^2 x$$ {$$Cot x=\frac{1}{Tan x}$$}
We know $$Tan x=\frac{Sin x}{cos x}$$
$$=\frac{Sin^2x}{Cos^2x}$$-------1
Given $$Sinx=\frac{4}{5}$$
cos x = base/hypotenuse
$$base^2=hypotenuse^2-Perpendicular^2$$
=$$5^2-4^2=25-16=9$$
base=3, Cos x=$$\frac{3}{5}$$
Put in equation one
$$=\frac{(\frac{4}{5})^2}{(\frac{3}{5})^2}$$
$$=\frac{16}{9}$$
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