In ΔABC if the angle A, B, C are in A.P. then a+ca2−ac+c2=

B=60o, b2=c2+a2−2caCosB B=60∘⇒b2=c2+a2−2cacos⁡B=c2+a2−caa+ca2−ac+c2=a+cb=sin⁡A+sin⁡Csin⁡B=2sin⁡[(A+C)/2]cos⁡[(A−C)/2]2sin⁡(B/2)cos⁡(B/2)=cos⁡[(A−C)/2]sin⁡30∘=2cos⁡A−C2.

In ΔABC if the angle A, B, C are in A.P. then a+ca2−ac+c2=

B=60o, b2=c2+a2−2caCosB B=60∘⇒b2=c2+a2−2cacos⁡B=c2+a2−caa+ca2−ac+c2=a+cb=sin⁡A+sin⁡Csin⁡B=2sin⁡[(A+C)/2]cos⁡[(A−C)/2]2sin⁡(B/2)cos⁡(B/2)=cos⁡[(A−C)/2]sin⁡30∘=2cos⁡A−C2.

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In $Δ A B C$ if the angle A, B, C are in A.P. then $\frac{a + c}{\sqrt{a^{2} - a c + c^{2}}} =$

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$C o s (\frac{A - C}{2})$

$2 C o s (\frac{A - C}{2})$

$2 S i n (\frac{A - C}{2})$

$2 C o s (\frac{A + C}{2})$

answer is B.

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Detailed Solution

$B = 60^{o}, b^{2} = c^{2} + a^{2} - 2 c a C o s B$

$B = 60^{\circ} \Rightarrow b^{2} = c^{2} + a^{2} - 2 cacos B = c^{2} + a^{2} - ca$
$\begin{array}{l} \frac{a + c}{\sqrt{a^{2} - ac + c^{2}}} = \frac{a + c}{b} = \frac{\sin A + \sin C}{\sin B} = \frac{2 \sin [(A + C) / 2] \cos [(A - C) / 2]}{2 \sin (B / 2) \cos (B / 2)} \\ = \frac{\cos [(A - C) / 2]}{\sin 30^{\circ}} = 2 \cos [\frac{A - C}{2}] . \end{array}$