This question was uploaded on 17/11/21 on social media accounts.
Solution:
A, B, C are angles of a triangle:
\[\Rightarrow A+B+C=180°\]
Given equation:
\[\cos A \cos B + \sin A \sin B \sin C = 1\]
Multiply both sides by 2:
\[2\cos A \cos B + 2\sin A \sin B \sin C = 2\]
Expand 2 or (1 + 1) by using:
\[\color{blue} {\sin^2\theta + \cos^2\theta = 1}\]
\[\Rightarrow (\sin^2A + \sin^2B - 2\sin A \sin B) + (\cos^2A + \cos^2B - 2\cos A \cos B)\]
\[= 2\sin A \sin B \sin C - 2\sin A \sin B\]
\[= 2\sin A \sin B (\sin C - 1)\]
\[\Rightarrow (\sin A - \sin B)^2 + (\cos A - \cos B)^2 + 2\sin A \sin B (1 - \sin C ) = 0\]
All three terms are positive (Explanation):
We can observe that the first two terms are positive due to square terms and in the third term, SinA and SinB and SinC are always positive due to angles of triangle and SinC is always less than 1 which means the third term is also positive.
Now, all terms sum is zero which means each term will be zero. Now we will get three equations for this:
\[\sin A = \sin B\]
\[\cos A = \cos B\]
\[\Rightarrow A = B\]
And the third equation:
\[2\sin A \sin B (1 - \sin C ) = 0\]
\[\Rightarrow 1 - \sin C = 0\]
\[\Rightarrow \sin C = 1\]
\[\Rightarrow C = 90°\]
By this we will get:
\[\Rightarrow A = B = 45°\]
By this: {A, B, C} = {45°, 45°, 90°}
That means the triangle is an isosceles right angled triangle.
