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 = \s

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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 $2 \sin^2 B = 2 \sin A \cos A = \sin 2A$. Since $\sin 2A \leq 1$,it follows that $2 \sin^2 B \leq 1$,or $\sin^2 B \leq \frac{1}{2}$. Now,for the quadratic equation $x^2 + 2x \cot B + 1 = 0$,the discriminant $D$ is given by: $D = (2 \cot B)^2 - 4(1)(1) = 4 \cot^2 B - 4 = 4(\cot^2 B - 1)$. Since $\sin^2 B \leq \frac{1}{2}$,we have $\csc^2 B \geq 2$,which implies $\cot^2 B = \csc^2 B - 1 \geq 2 - 1 = 1$. Therefore,$\cot^2 B - 1 \geq 0$,which means $D \geq 0$. Since the discriminant is non-negative,the roots are always real.

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

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