Question

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

• A. Real and distinct
• B. real
• C. imaginary
• D. Real and Equal

Answer 1

Answer: (B)

real

${\sin }^{2}B=\cos A\sin A$

$\Rightarrow {\sin }^{2}B=\frac{\sin 2A}{2}$

$\Rightarrow {\text{cosec}}^{2}B=\frac{2}{\sin 2A}$

$\Rightarrow {\cot }^{2}B=\frac{2}{\sin 2A}-1$

Now, $\frac{\sin 2A}{2}\ge 0$ because ${\sin }^{2}B\ge 0$

And $\sin 2A\le 1$

$\Rightarrow \frac{2}{\sin 2A}\ge 2,{\cot }^{2}B\ge 1$

Dicriminant $=b^2-4ac=4({\cot }^{2}B-1)$

$\Rightarrow 4({\cot }^{2}B-1)\ge 0$

Hence real root or roots.