The key point to solve the problem:

The idea of basic trigonometric formulae, i.e. transformation and T – ratios of multiple angles

Given,

cos²A + cos² B + cos² C = 1

Multiplying 2 to both sides so that we can change it in Trigonometric ratios of multiple angles so that we can get the value of angle.

As 2 cos²X = 1 + cos 2X

∴ 2cos²A + 2cos² B + 2cos² C = 2

⇒ 1 + cos 2A + 1 + cos 2B + 1 + cos 2C = 2

⇒ cos 2A + cos 2B + cos 2C = -1

Using, cos 2X + cos 2Y = 2 cos (X + Y) cos (X – Y)

⇒ 2 cos (A + B) cos (A – B) = -1(1 + cos 2C)

As 2 cos²X = 1 + cos 2X and A + B + C = π

We have,

2 cos (π - C) cos (A – B) = -2 cos² C

-2 cos C cos (A – B) = -2 cos² C {∵ cos (π - θ) = - cos θ }

∴ 2 cos C ( cos C + cos (A – B)) = 0

Either cos C = 0 ⇒ ∠ C = 90°

Or cos C = -cos (A – B) ⇒ C = π – (A – B) which is not possible as in ΔABC, A + B + C = π

∴ C = 90° is the only satisfied solution.

Hence, Δ ABC is a right triangle, right angled at ∠ C …proved