Question

If $\sin A , \sin B , \cos A$ are in GP, then roots of $x ^{2}+2 x \cot B +1=0$ are always

• A. real
• B. imaginary
• C. greater than 1
• D. equal

Answer 1

The Correct Option is A

$\sin A, \sin B, \cos A$ are in G.P. $\therefore \sin ^{2} B =\sin A \cos A \rightarrow 1$ $x^{2}+2 x \cot B+1=0$ $D =(2 \cot B )^{2}-4(1)(1)$ $D =4 \cot ^{2} B -4$ $D=4\left[\frac{\cos ^{2} B}{\sin ^{2} B}-1\right]$ $D=4\left[\frac{\cos ^{2} B-\sin ^{2} B}{\sin ^{2} B}\right]$ $D=4\left[\frac{1-\sin ^{2} B-\sin ^{2} B}{\sin ^{2} B}\right] \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$ $D=4\left[\frac{1-2 \sin ^{2} B}{\sin ^{2} B}\right]$ $D=4\left[\frac{1-2 \sin A \cos A}{\sin ^{2} B}\right]$ (From equation 1) $D=4\left[\frac{\sin ^{2} A+\cos ^{2} A-2 \sin A \cos A}{\sin ^{2} B}\right]$ $D=4\left[\frac{(\sin A-\cos A)^{2}}{\sin ^{2} B}\right]$ $D=\left[\frac{2(\sin A-\cos A)}{\sin B}\right]^{2} \geq 0$ Therefore, roots of given equation are real.