We are given the expression cotxcot2x - cot2xcot3x - cot3xcotx and we need to prove that it is equal to 1.
We know that cot θ = cosθ/sinθ. Therefore, we can rewrite the expression in terms of sine and cosine:
(cosx/sinx)(cos2x/sin2x) - (cos2x/sin2x)(cos3x/sin3x) - (cos3x/sin3x)(cosx/sinx) = 1
Let's simplify this expression. We can find a common denominator:
(cosxcos2xsin3x - cos2xcos3xsinx - cos3xcosxsin2x) / (sinxsin2xsin3x)
We will use the trigonometric identities:
sin2θ = 2sinθcosθ
cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ
Using the triple angle formula for sine and cosine:
sin3x = 3sinx - 4sin³x
cos3x = 4cos³x - 3cosx
However, a simpler approach is to use the cotangent addition formula:
cot(A-B) = (cotAcotB + 1) / (cotB - cotA)
Let's rewrite the expression in a form that allows us to use this formula. Notice that the expression is similar to an expansion of the form cot(A-B) = cotAcotB + 1 / (cotB - cotA).
Let's consider cot(3x - x) = cot(2x).
Using the cotangent addition formula: cot(3x - x) = (cot3xcotx + 1) / (cotx - cot3x) = cot2x
Therefore, cot2x(cotx - cot3x) = cot3xcotx + 1
Rearranging the terms:
cot3xcotx - cot2xcotx + cot2xcot3x = -1
-cot3xcotx + cot2xcotx - cot2xcot3x = 1
cotxcot2x - cot2xcot3x - cot3xcotx = 1
Hence, the given expression is proven to be equal to 1.