Question

STATEMENT 1: In a $\Delta ABC,c{\cos }^2\frac{A}{2}+a{\cos }^2\frac{C}{2}=\frac{3b}{2}$ then the sides $a,b,c$ are in $A.P$

STATEMENT 2: In a $\Delta ABC,a\cos C+ccosA=b$

• A. Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement- 1
• B. Statement-1 is true, Statement-2 is true,Statement-2 is not correct explanation for Statement-1
• C. Statement-1is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

The correct option is A Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement- 1

According to Projection Rule,

$a.cosC+c.cosA=b$

Hence $Statement2$ is correct

we also know that $cosA=2co{s}^2\frac{A}{2}-1,cosC=2co{s}^2\frac{C}{2}-1$

$acosC+ccosA=b$ $\Rightarrow 2aco{s}^2\frac{C}{2}-a+2cco{s}^2\frac{A}{2}-c=b\Rightarrow 2aco{s}^2\frac{C}{2}+2cco{s}^2\frac{A}{2}=a+b+c\Rightarrow aco{s}^2\frac{C}{2}+cco{s}^2\frac{A}{2}=\frac{a+b+c}{2}$

Since, a,b,c are in $AP$

$a+c=2b$

so, we can write $\frac{a+b+c}{2}=\frac{3b}{2}$

Therefore,

$aco{s}^2\frac{C}{2}+cco{s}^2\frac{A}{2}=\frac{3b}{2}$

$Statement1$ is also correct and is correctly explained by $Statement2$

Therefore, Correct Answer is $A$