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

Verify the identity

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Howard L. | Experienced Math Tutor Specializing in Test Prep & Grade Up SkillsExperienced Math Tutor Specializing in T...
5.0 5.0 (21 lesson ratings) (21)
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.
Sanju S. | Elementary to College Level Math TutoringElementary to College Level Math Tutorin...
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 cos2
       = (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])
         = [(1/ sin2x) cos2x] - cos2x
         =  cos2x/ sin2x - cos2x
cosx/ sinx = cot x
then, cos2x/sin2x = cot2x
  cos2x/ sin2x - cos2x = cot2x - cos2x = LHS
Francisco E. | Francisco; Civil Engineering, Math., Science, Spanish, Computers.Francisco; Civil Engineering, Math., Sci...
5.0 5.0 (1 lesson ratings) (1)
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)


Brenda, please tell me what you do not understand, is the process? is the different replacements? I can go as detailed as you ask me for.
Howard, I do not say that your answer is wrong but I learned that one of the sides need to be left as in the original problem, because they are asking for making one side equal to the original other. I liked your solution very much but i learned it in different way. No polemics.