Question

Consider a triangle ABC satisfying $2a{\sin }^2\left(\frac{C}{2}\right)+2c{\sin }^2\left(\frac{A}{2}\right)=2a+2c-3b$, then
$\sin A,\sin B,\sin C$ are in

• A. G.P.
• B. A.P.
• C. H.P.
• D. Neither in G.P. nor in A.P. nor in H.P.

Answer 1

Answer: (B)

A.P.

In a $△ABC,$

$2a{\sin }^2\frac{C}{2}+2c{\sin }^2\frac{A}{2}=2a+2c-3b$

or, $a(1-\cos C)+c(1-\cos A)=2a+2c-3b$ ...... $[\because \cos 2x=1-2{\sin }^{2}x]$

or, $a+c-(\cos C+\cos A)=2a+2c-3b$ (Properties of traingles)

or, $a+c-b=2a+2c-3b$

or, $a+c=2b$

$\therefore a,b,c$ are in A.P.

Since, $a,b$ and $c$ are in $A.P.$

We know that,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=K$

Also, If each term of an $A.P$ is multiplied or divided by a non-zero constant $K,$ then the resulting sequence is also in $AP.$

So, $K\sin A,K\sin B$ and $K\sin C$ are also in $A.P.$

Hence, $\sin A,\sin B,\sin C$ are in $A.P.$

Hence, B is the correct option.