Answer :


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


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