Observe the following statements (I) In $\triangle A B C, b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}=s$…
Observe the following statements
(I) In $\triangle A B C, b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}=s$
(II) In $\triangle A B C, \cot \frac{A}{2}=\frac{b+c}{2} \Rightarrow B=90^{\circ}$
Which of the following is correct?
Both I and II are true.
I is true, II is false.
I is false, II is true.
Both I and II are false.
Solution
(I) Now,
$
\begin{aligned}
& , b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2} \\
& =b \cdot \frac{s(s-c)}{a b}+c \cdot \frac{s(s-b)}{a c} \\
& =\frac{s}{a}(s-c+s-b) \\
& =\frac{s}{a}(2 s-(b+c))=\frac{s}{a} \cdot a \\
& =s
\end{aligned}
$
$\therefore$ Statement Ist is true.
$
\begin{array}{rlrl}
\text { (II) Let } \cot \frac{A}{2} & =\frac{b+c}{a} \\
& \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} & =\frac{\sin B+\sin C}{\sin A} \\
\Rightarrow \quad & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} & =\frac{2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)}{2 \sin \frac{A}{2} \cos \frac{A}{2}} \\
\Rightarrow & \cos \frac{A}{2} & =\cos \left(\frac{B-C}{2}\right) \\
\Rightarrow & \frac{A}{2} & =\frac{B-C}{2} \\
\Rightarrow & A+C & =B
\end{array}
$
But $A+B+C=\pi$, therefore $B=\frac{\pi}{2}$
But given statement is
$
\cot \frac{A}{2}=\frac{b+c}{2} \Rightarrow B=90^{\circ} .
$
$\therefore$ Statement-II is not true.
Hence, option (2) is correct