Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $(\sin A+\sin B+\sin C)(\sin A+\sin C-\sin B)=\mu \sin A\sin C$,

where $\sin A=ak,\sin B=bk,\sin C=ck$

then the range of $\mu$ is

• A. $[0,1]$
• B. $[-4,4]$
• C. $(0,4)$
• D. $[0,4]$

Answer 1

The correct option is C $(0,4)$

$(\sin A+\sin C-\sin B)(\sin A+\sin C+\sin B)$

$=(\sin A+\sin C{)}^2-si{n}^{2}B$

$=\mu \sin A.\sin C$

Now

$a=k\sin A$ , $b=k\sin B$ and $c=k\sin C$.

Hence

$(a+c{)}^2-b^2=\mu ac$

Or

$(a^2+c^2-b^2)+2ac=\mu ac$

Or

$2ac[\frac{a^2+c^2-b^2}{2ac}+1]=\mu ac$

$2ac.(\cos B+1)=\mu ac$

Or

$2(\cos B+1)=\mu$

Or

$4{\cos }^2\left(\frac{B}{2}\right)=\mu$

Hence

$\mu ϵ(0,4)$...................$co{s}^2\left(\theta \right)\in [0,1]$