Question

Given that the lengths of three sides $BC,CA,AB$ of $△ABC$ are $a,b,c$ respectively and $a,b,c$ are in geometric progression. Then, the value of the expression $\frac{\sin A\cot C+\cos A}{\sin B\cot C+\cos B}$ lies in

• A. $(0,\frac{\sqrt{5}-1}{2})$
• B. $(0,\frac{\sqrt{5}+1}{2})$
• C. $(\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}+1}{2})$
• D. $(\frac{\sqrt{5}-1}{2},2)$

Answer 1

The correct option is C $(\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}+1}{2})$

Given, $a,b,c$ are in G.P.
Let $b=ar,c=a{r}^2$
where $r$ is the common ratio of G.P.

Let $E=\frac{\sin A\cot C+\cos A}{\sin B\cot C+\cos B}$
$\Rightarrow E=\frac{\sin A\cos C+\cos A\sin C}{\sin B\cos C+\cos B\sin C}$
$\Rightarrow E=\frac{\sin (A+C)}{\sin (B+C)}$
$\Rightarrow E=\frac{\sin B}{\sin A}\Rightarrow E=\frac{b}{a}\Rightarrow E=r$

Now using triangle inequality,
$a+b>c\Rightarrow a+ar>a{r}^2$
$\Rightarrow r^2-r-1<0$
$\Rightarrow 0<r<\frac{\sqrt{5}+1}{2}\dots \left(1\right)$
(as $r$ is positive)

and $b+c>a$
$\Rightarrow ar+a{r}^2>a\Rightarrow r^2+r-1>0$
$\Rightarrow r>\frac{\sqrt{5}-1}{2}\dots \left(2\right)$

Here, there is no need to take the third inequality because either $a$ or $c$ will be the greatest side of triangle.

From $\left(1\right)\cap \left(2\right),$ we get
$r\in (\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}+1}{2})$