Question

If $\sin A,\sin B$ and $\cos A$ are in GP, then the 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

Explanation for the correct option:

Roots of an quadratic equation:

Given that, $\sin A,\sin B$ and $\cos A$ are in G.P.

Then, by definition of G.P which states that if $a,b,c$are in G.P then $b^2=ac$.

Here, $a=\sin A,b=\sin B,c=\cos A$.

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

Now using discriminant, $D=b^2-4ac$ we determine the nature of roots.

We have equation $x^2+2x\cot B+1=0$ .

$\begin{array}{rcl}& \Rightarrow & D=4co{t}^{2}B-4\\ & =& 4\left(co{t}^{2}B-1\right)\\ & =& 4\left(\cos e{c}^{2}B-2\right)>0\mathbf{[}\mathbf{\because }\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}{\mathbf{t}}^{\mathbf{2}}\mathbf{B}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{B}\mathbf{]}\\ & =& 4\frac{(1-2{\sin }^{2}B)}{{\sin }^{2}B}\\ & =& 4\frac{({\sin }^{2}A+{\cos }^{2}A-2\sin A\cos A)}{{\sin }^{2}B}\mathbf{}\left[\because si{n}^{2}A+co{s}^{2}A=1,si{n}^{2}B=sinAcosA\right]\\ & =& 4\frac{{(\sin A-\cos A)}^2}{{\sin }^{2}B}\end{array}$

Since discriminant is $>0$ so the equation has real roots.

Hence, option A is correct.