Question

If in a $\Delta ABC$ angles $A,B,C$ are in A.P then $\frac{a+c}{\sqrt{a^2-ac+c^2}}=$

• A. $2\sin \frac{A-C}{2}$
• B. $\sin \frac{A-C}{2}$
• C. $2\cos \frac{A-C}{2}$
• D. $\cos \frac{A-C}{2}$

Answer 1

Answer: (C)

$2\cos \frac{A-C}{2}$

Given angles $A,B,C$ are in A.P

let $A=P-DB=PC=P+D$

$A+B+C=\pi \Rightarrow P-D+P+P+D=\pi \Rightarrow 3P=\pi \Rightarrow P=\frac{\pi }{3}B=\frac{\pi }{3}$

from sine rule of triangle $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$=k$

from cosine rule of triangle $\cos B=\frac{c^2{+a}^2{-b}^2}{2ac}\Rightarrow \cos \frac{\pi }{3}=\frac{c^2{+a}^2{-b}^2}{2ac}\Rightarrow \frac{1}{2}=\frac{c^2{+a}^2{-b}^2}{2ac}\Rightarrow b^2=a^2-ac+c^2$

$\frac{a+c}{\sqrt{a^2-ac+c^2}}=\frac{a+c}{\sqrt{b^2}}=\frac{a+c}{b}=\frac{k\sin A+k\sin C}{k\sin B}=\frac{\sin (P-D)+\sin (P+D)}{\sin P}=\frac{\sin P\cos D-\sin D\cos P+\sin P\cos D+\sin D\cos P}{\sin P}=\frac{2\sin P\cos D}{\sin P}=2\cos D=2\cos -D=2\cos \left(\frac{(P-D)-(P+D)}{2}\right)=2\cos \left(\frac{A-C}{2}\right)$

so option is C is correct