Verify the identity

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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...

cot^{2}x-cos^{2}x=cot^{2}x cos^{2}x

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

Trig Identity: 1 + cot^{2}x = csc^{2}x

cot^{2}x = csc^{2}x - 1 (substitute this on RHS)

RHS = cot^{2}x cos^{2}x

= (csc^{2}x - 1) cos^{2}x (now use distributive property to expand)

= csc^{2}x* cos^{2}x - cos^{2}x [1]

trig identityL

csc x = 1/sinx

therefore, csc^{2}x = 1/ sin^{2}x (substitute in [1])

Gives:

= [(1/ sin^{2}x) cos^{2}x] - cos^{2}x

= cos^{2}x/ sin^{2}x - cos^{2}x

but,

cosx/ sinx = cot x

then, cos^{2}x/sin^{2}x = cot^{2}x

thus,

cos^{2}x/ sin^{2}x - cos^{2}x = cot^{2}x - cos^{2}x = LHS

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)

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