Question

In a triangle $ABC,s-a,s-b,s-c$ are in GP, then $\frac{{\sin }^{2}A+{\sin }^{2}C}{\sin A+\sin C}$ is equal to

• A. $\sin \left(A+C\right)$
• B. $\sin \left(A-C\right)$
• C. $\frac{b}{R}$
• D. None of these

Answer 1

The correct option is A

$\sin \left(A+C\right)$

Explanation for the correct option:

Step- 1: Form an equation.

As the terms $s-a,s-b,s-c$ are in GP, so

$$\begin{gathered} {\left(s-b\right)}^2=\left(s-a\right)\left(s-c\right) \\ \Rightarrow s^2-2bs+b^2=s^2-sc-sa+ac \\ \Rightarrow \left(a+c-2b\right)s=ac-b^2 \\ \Rightarrow \left(a+c-2b\right)\left(\frac{a+b+c}{2}\right)=ac-b^2 \\ \Rightarrow a^2+ab+ac+ac+bc+c^2-2ab-2b^2-2bc=2ac-2b^2 \\ \Rightarrow a^2+c^2=ab+bc \\ \Rightarrow \frac{a^2+c^2}{a+c}=b \end{gathered}$$

Step 2. Find the value of $\frac{{\sin }^{2}A+{\sin }^{2}C}{\sin A+\sin C}$.

Using the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\left(\mathrm{let}\right)$ the equation $\frac{a^2+c^2}{a-c}=b$ can be written as:

$$\begin{gathered} \frac{{\left(k\sin A\right)}^2+{\left(k\sin C\right)}^2}{k\sin A-k\sin C}=k\sin B \\ \Rightarrow \frac{k^2\left({\sin }^{2}A+{\sin }^{2}C\right)}{k\left(\sin A+\sin C\right)}=k\sin B \\ \Rightarrow \frac{\left({\sin }^{2}A+{\sin }^{2}C\right)}{\left(\sin A+\sin C\right)}=\sin B \\ \Rightarrow \frac{\left({\sin }^{2}A+{\sin }^{2}C\right)}{\left(\sin A+\sin C\right)}=\sin \left(180°-\left(A+C\right)\right)\left[\mathit{A}\mathbf{+}\mathit{B}\mathbf{+}\mathit{C}\mathbf{=}\mathbf{180}\mathbf{°}\right] \\ \Rightarrow \frac{\left({\sin }^{2}A+{\sin }^{2}C\right)}{\left(\sin A+\sin C\right)}=\sin \left(A+C\right)\left[\mathbf{sin}\mathbf{(}\mathbf{180}\mathbf{°}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{sin}\mathit{\theta }\right] \end{gathered}$$

So the value of $\frac{{\sin }^{2}A+{\sin }^{2}C}{\sin A+\sin C}$ is $\sin \left(A+C\right)$.

Hence, the correct option is A.