Question

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

• A. Real
• B. Imaginary
• C. Greater than $1$
• D. Equal

Answer 1

The correct option is A Real

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