Question 66

If $$\sin x = \frac{4}{5}$$, then $$\frac{Tan x}{Cot x}$$ = ?

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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