Question
In a $$\vartriangle ABC,\cos B \cdot \cos C + \sin B \cdot \sin C \cdot {\sin ^2}A = 1.$$ Then the triangle is
A.
right-angled isosceles
B.
isosceles whose equal angles are greater than $$\frac{\pi }{4}$$
C.
equilateral
D.
None of these
Answer :
right-angled isosceles
Solution :
$$\cos B \cdot \cos C + \sin B \cdot \sin C \geqslant 1$$
because $$\sin B \cdot \sin C \cdot {\sin ^2}A$$ is positive and $${\sin ^2}A \leqslant 1$$
or, $$\cos \left( {B - C} \right) \geqslant 1.\,{\text{But}}\,\,\cos \left( {B - C} \right) > 1.\,{\text{So,}}\,\,\cos \left( {B - C} \right) = 1.$$
Therefore, $$B = C$$ and then $$\sin A = 1.$$
$$\therefore \,\,A = \frac{\pi }{2},B = C = \frac{\pi }{4}.$$