Question

In $\Delta ABC$ if the angle $A,B,C$ are in $A.P$. then $\frac{a+c}{\sqrt{a^2-ac+c^2}}$ is equal to

• A. $\cos \left(\frac{A-C}{2}\right)$
• B. $2\cos \left(\frac{A-C}{2}\right)$
• C. $2\sin \left(\frac{A-C}{2}\right)$
• D. $2\cos \left(\frac{A+C}{2}\right)$

Answer 1

Answer: (B)

$2\cos \left(\frac{A-C}{2}\right)$

In given triangle, ABC, angles A,B,C are in A.P.

$\Rightarrow$ $2B=A+C$

Also, $A+B+C=\pi$

Therefore, $\frac{A+C}{2}=B=\frac{\pi }{3}$

By cosine formula,

$\cos B=\frac{a^2=+c^2-b^2}{2ac}$

Putting value of B, we get,

$a^2+c^2-ac=b^2$ .... $e{q}^n\left(1\right)$

By sine formula,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ ..... $e{q}^n\left(2\right)$

Therefore, $\frac{a+c}{\sqrt{a^2-ac+c^2}}=\frac{2R(\sin A+\sin C)}{b}$ ..... [from $e{q}^n\left(1\right),\left(2\right)$]

$=\frac{2R(\sin A+\sin C)}{2R\left(\sin B\right)}$

$=\frac{2\times \sin \left(\frac{A+C}{2}\right)\times \cos \left(\frac{A-C}{2}\right)}{\sin B}$

$=2\times \cos \left(\frac{A-C}{2}\right)$