cot^2x-cos^2x=cot^2x cos^2x

Question

Verify the identity

Answer 1

Here is another way to verify by working on both side together,

cot^2x - cos^2x = cot^2x * cos^2x,  where cot^2x = (cos^2x)/(sin^2x)

Substitute cot^2x and multiply the second term with (sin^2x)/(sin^2x) for common denominator

(cos^2x)/(sin^2x) -( cos^2x)(sin^2x)/(sin^2x) = (cos^2x)/(sin^2x) * cos^2x

Multiply by sin^2x on both sides to eliminate denominator and factor, then

cos^2x(1 - sin^2x) = cos^2x *  cos^2x

Devide by cos^2x on both sides

1- sin^2x = cos^2x,  where  1 - sin^2x = cos^2x

Then, cos^2x = cos^2x,  both sides same.

 

Answer 2

cot²x-cos²x=cot²x cos²x

 

Work with Right Hand Side to proof it's equivalent to L.H.S.

 

Trig Identity: 1 + cot²x = csc²x

                  

                   cot²x = csc²x - 1 (substitute this on RHS)

 

RHS = cot²x cos²x 

 

       = (csc²x - 1) cos²x        (now use distributive property to expand)

       = csc²x* cos²x - cos²x     [1]

 

trig identityL

 

csc x = 1/sinx

 

therefore, csc²x = 1/ sin²x (substitute in [1])

 

Gives:

         = [(1/ sin²x) cos²x] - cos²x

         =  cos²x/ sin²x - cos²x

 

but, 

cosx/ sinx = cot x

then, cos²x/sin²x = cot²x

 

thus, 

  cos²x/ sin²x - cos²x = cot²x - cos²x = LHS

       

Answer 3

We have the equality:

Cot^2 X - cos^2 X = Cot^2 X * Cos^2 X

I will work with the left hand side of the equality

(Cos^2 X/Sin^2 X) - Cos^2 X =...........

making the common denominator

(Cos^2 X - Cos^2 X * Sin^2 X)/ sin^2 X , Factorizing Cos^2 X

Cos^2 X ( 1 - Sin^2 X)/ sin^2 X =..........

Cos^2 X/ Sin^2 X = Cot^2 X

Cot^2 X (1- Sin^2X) = ...........

But (1-Sin^2 X) = (Sin^2 X + Cos^2 X - Sin^2 X) = Cos^2 X

So the left side will turn into Cot^2 X * Cos^2 X which is exactly the same right hand part of the equality.

This was what you were asking to verify. CHECK!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I used two common equalities: (Sin^2 X + Cos^2 X = 1) and ( Cot^2X = Cos^ X/ Sin^2 X)