Verify identity cosxcotx/cotx-cosx=cotx+cosx/cosxcotx

Question

able to do the first 4 steps and the stuck-

Answer 1

cos(x)cot(x)/(cot(x) - cos(x)) = (cot(x) + cos(x))/cos(x)cot(x)

Multiply both sides by cos(x)cot(x)

cos²(x)cot²(x)/(cot(x) - cos(x)) = cot(x) + cos(x)

Multiply both sides by (cot(x) - cos(x))

cos²(x)cot²(x) = (cot(x) + cos(x))(cot(x) - cos(x))

cos²(x)cot²(x) = cot²(x) - cos²(x)

Add cos²(x) to both sides

cos²(x)cot²(x) + cos²(x) = cot²(x)

Factor out cos²(x) on left side

cos²(x)(cot²(x) + 1) = cot²(x)

Using the identity (cot²(x) + 1) = csc²(x)

cos²(x)(csc²(x)) = cot2(x)

Use the identity csc²(x) = (1/sin²(x))

cos²(x)(1/sin²(x)) = cot²(x)

cos²(x)/sin²(x) = cot²(x)

Take sqrt both sides

cos(x)/sin(x) = cot(x)

Answer 2

first cross multiply to get. (Cos^2)(Cot^2)= (Cot^2)-(Cos^2). Next divide both sides by (Cot^2)which gives you

(Cos^2) = 1-(Cos^2)/(Cot^2)

Now let’s just work on the last term and substitute cos^2/sin^2 for cot^2

cos^2/(cos^2/sin^2)=sin^2 so we have

Cos^2= 1-Sin^2 and 1-sin^2 does equal cos^2