If in a Delta ABC,cos A cos B + sin A sin B s | vedclass

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Question

(A) Given $\cos A \cos B + \sin A \sin B \sin^2 C = 1$.Since $\sin^2 C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin^2 C \le \cos A \cos B + \sin

Vedclass · Accepted Answer

$\Delta ABC$ is isosceles but not right-angled

Vedclass · Answer

(A) Given $\cos A \cos B + \sin A \sin B \sin^2 C = 1$.Since $\sin^2 C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin^2 C \le \cos A \cos B + \sin A \sin B = \cos(A - B)$.Thus,$\cos(A - B) \ge 1$,which implies $\cos(A - B) = 1$,so $A = B$.Substituting $A = B$ into the original equation: $\cos^2 A + \sin^2 A \sin^2 C = 1$.$\sin^2 A \sin^2 C = 1 - \cos^2 A = \sin^2 A$.Since $A$ is an angle of a triangle,$\sin A 
eq 0$,so $\sin^2 C = 1$,which means $C = \frac{\pi}{2}$.Since $A + B + C = \pi$ and $A = B$,we have $2A + \frac{\pi}{2} = \pi$,so $A = B = \frac{\pi}{4}$.Therefore,the triangle is an isosceles right-angled triangle with angles $(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2})$.Statement $(A)$ is incorrect because the triangle is right-angled.

If in a $\Delta ABC$,$\cos A \cos B + \sin A \sin B \sin^2 C = 1$,then the statement which is incorrect is:

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