Explanation:

Let ABC be a triangle and let a, b, c be the lengths of the sides opposite to the angles A, B and C respectively.

Given, a:b = cos C : cos B, b:c = cos A : cos C, c:a = cos B : cos A

Step 1: Simplify the given equation

We know that cos(A) = (b² + c² - a²)/(2bc), cos(B) = (c² + a² - b²)/(2ca) and cos(C) = (a² + b² - c²)/(2ab).

By substituting these values in the given equations, we get

a² = b² + c² - 2bc(cos A/cos C)

b² = c² + a² - 2ca(cos B/cos A)

c² = a² + b² - 2ab(cos C/cos B)

Step 2: Prove that a = b = c

We need to prove that a = b = c for the triangle to be equilateral.

We know that the sum of the squares of all three sides of a triangle is equal to twice the sum of the squares of its altitudes.

So,

a² + b² + c² = 2(ha² + hb² + hc²)

where ha, hb and hc are the altitudes of the triangle.

Now, substituting the values of a², b² and c², we get

3(a² + b² + c²) = 2(bc(cos A/cos C) + ca(cos B/cos A) + ab(cos C/cos B))

= 2abc(cos A/cos C + cos B/cos A + cos C/cos B)

= 2abc(cos² A cos B + cos² B cos C + cos² C cos A)/(cos A cos B cos C)

= 2abc(cos A cos B cos C)/(cos A cos B cos C)

= 2abc

Therefore, a² + b² + c² = 2abc

Now, if we assume that the triangle is not equilateral, then at least two sides of the triangle are different. Without loss of generality, let a be the longest side.

Then, b + c > a

Squaring both sides, we get

b² + 2bc + c² > a²

Substituting the value of a² from the equation derived in Step 1, we get

2bc(cos A/cos C) + 2ca(cos B/cos A) + 2ab(cos C/cos B) > 4bc(cos² A/cos² C)

+ 4ca(cos² B/cos² A) + 4ab(cos² C/cos² B)

= 4abc(cos² A cos B + cos² B cos C + cos² C cos A)/(cos² A cos² B cos² C)

But we know that cos A cos B cos C ≤ (3√3)/8 and cos² A cos B + cos² B cos C + cos² C cos A ≤ 3/4.

Therefore,

2bc(cos A/cos

Solution:

Given, in a triangle ABC, the sides are proportional to the cosines of the opposite angles.

Let AB = k cos C, BC = k cos A and AC = k cos B.

We know that in a triangle, the sum of the cosines of the angles is equal to 1.

Hence, cos A + cos B + cos C = 1.

Substituting the values of the sides in the above equation, we get:

(cos A + cos B + cos C) = cos A + cos B + cos (180° - A - B)
= cos A + cos B - cos A cos B + sin A sin B
= (cos A + sin A sin B) + (cos B + sin A sin B) - cos A cos B
= (cos A + sin A sin B) + (cos B + sin B sin A) - cos A cos B
= (cos A + cos B) + sin A sin B - cos A cos B
= 1 + sin A sin B - cos A cos B
= 1 + (sin A sin B + cos A cos B) - cos A cos B
= 1 + sin (A + B) - cos A cos B

Since A + B + C = 180°, we have A + B = 180° - C.

Substituting this in the above equation, we get:

cos A + cos B + cos C = 1 + sin C - cos A cos B

cos C = 1 + sin C - cos A cos B - cos A - cos B
cos C - cos A - cos B = sin C - cos A cos B - 1

Squaring both sides, we get:

(cos C - cos A - cos B)² = (sin C - cos A cos B - 1)²

Simplifying, we get:

cos² A + cos² B + cos² C - 2 cos A cos B - 2 cos B cos C - 2 cos C cos A = sin² C + cos² A cos² B + 1 - 2 sin C cos A cos B

Since cos² A + cos² B + cos² C = 1 + 2 cos A cos B cos C, we have:

1 + cos A cos B cos C - cos A cos B - cos B cos C - cos C cos A = sin² C + cos² A cos² B + 1 - 2 sin C cos A cos B

1 + cos A cos B cos C = sin² C + cos² A cos² B + 2 sin C cos A cos B
1 + cos A cos B cos C = 1 - cos² C + cos² A cos² B + 2 sin C cos A cos B
cos² C + cos A cos B cos C = cos² A cos² B + 2 sin C cos A cos B
cos C (cos A cos B - cos C) = cos² A cos² B - sin² C

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Hence, cos² A cos² B - sin² C < />

Since sin² C is non-negative, we