Question

If $2b=(m+1)a$ and $\cos A=\frac{1}{2}\sqrt{\left(\frac{(m-1)(m+3)}{m}\right)},$ where $1<m<3,$ then $\frac{c_1}{c_2}$ is

• A. $m$ or $\frac{1}{m}$
• B. $(m-1)$ or $\frac{1}{(m-1)}$
• C. $(m+1)$ or $\frac{1}{(m+1)}$
• D. $(m+3)$ or $\frac{1}{(m+3)}$

Answer 1

The correct option is A $m$ or $\frac{1}{m}$

$\because \cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow c^2-2bc\cos A+b^2-a^2=0$
$\Rightarrow c^2-2(m+1)\lambda c\cos A+(m^2+2m-3){\lambda }^2=0$
$\because \frac{a}{2}=\frac{b}{m+1}=\lambda$(say)
$\therefore c_1+c_2=2(m+1)\lambda \cos A$
or ${(c_1+c_2)}^2=4{(m+1)}^2{\lambda }^2\frac{(m-1)(m+3)}{4m}$
$=\frac{{(m+1)}^2(m-1)(m+3)}{4}$ .................$\left(1\right)$
and ${(c_1-c_2)}^2={(c_1+c_2)}^2-4c_1{c}_2$
$={4(m+1)}^2{\lambda }^2{\cos }^{2}A-4(m^2+2m-3){\lambda }^2$
$=4{\lambda }^2[\frac{{(m+1)}^2(m-1)(m+3)}{4m}-(m+3)(m-1)]$
$=4{\lambda }^2\frac{(m-1)(m+3){(m-1)}^2}{4m}$
$=\frac{{\lambda }^2{(m-1)}^3(m+3)}{m}$ .............$\left(2\right)$
From eqns$\left(1\right)$ and $\left(2\right)$
${\left(\frac{c_1+c_2}{c_1-c_2}\right)}^2$
$=\left(\frac{{(m+1)}^2(m-1)(m+3)}{m}\right)\times \left(\frac{m}{{\lambda }^2{(m-1)}^3(m+3)}\right)$
$=\frac{{(m+1)}^2}{{(m-1)}^2}$
$\Rightarrow \frac{c_1+c_2}{c_1-c_2}=\frac{m+1}{m-1}$ or $\frac{m+1}{1-m}$
At last using the componendo-dividendo rule, we get
$\frac{c_1+c_2-c_1+c_2}{c_1+c_2+c_1-c_2}=\frac{m+1-m+1}{m+1+m-1}$ or $\frac{m+1-1+m}{m+1+1-m}$
$\frac{2c_1}{2c_2}=\frac{2}{2m}$ or $\frac{2m}{2}$
$\therefore \frac{c_1}{c_2}=\frac{1}{m}$ or $m$