If sin A, sin B, cos A are in G.P.,then the r | vedclass

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(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.Multiplying by $2$,we get $2 \sin^2 B = 2 \sin A \cos A = \si

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(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.Multiplying by $2$,we get $2 \sin^2 B = 2 \sin A \cos A = \sin 2A$.Using the identity $2 \sin^2 B = 1 - \cos 2B$,we have $1 - \cos 2B = \sin 2A$,which implies $\cos 2B = 1 - \sin 2A$.Since $-1 \le \sin 2A \le 1$,it follows that $0 \le 1 - \sin 2A \le 2$,so $\cos 2B \ge 0$.For the quadratic equation $x^2 + 2x \cot B + 1 = 0$,the discriminant $D$ is given by $D = (2 \cot B)^2 - 4(1)(1) = 4 \cot^2 B - 4 = 4(\cot^2 B - 1)$.Using $\cot^2 B - 1 = \frac{\cos^2 B - \sin^2 B}{\sin^2 B} = \frac{\cos 2B}{\sin^2 B}$,we get $D = 4 \frac{\cos 2B}{\sin^2 B}$.Since $\cos 2B \ge 0$ and $\sin^2 B > 0$,we have $D \ge 0$.Therefore,the roots of the equation are always real.

If $\sin A, \sin B, \cos A$ are in $G.P.$,then the roots of the equation $x^2 + 2x \cot B + 1 = 0$ are always

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