Observe the following statements:(I) In trian | vedclass

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(B) For statement $(I)$:$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = b \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-b)}{ac}$$= \frac{s}{a}(s-c) + \fr

Vedclass · Accepted Answer

$I$ is true,$II$ is false.

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(B) For statement $(I)$:$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = b \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-b)}{ac}$$= \frac{s}{a}(s-c) + \frac{s}{a}(s-b) = \frac{s}{a}(2s - b - c)$Since $2s = a+b+c$,we have $2s - (b+c) = a$.So,$\frac{s}{a} \cdot a = s$. Thus,statement $(I)$ is true.For statement $(II)$:Given $\cot \frac{A}{2} = \frac{b+c}{a}$. Using sine rule,$\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A}$.$\frac{\cos(A/2)}{\sin(A/2)} = \frac{2 \sin((B+C)/2) \cos((B-C)/2)}{2 \sin(A/2) \cos(A/2)}$.Since $\sin((B+C)/2) = \cos(A/2)$,we get $\cos(A/2) = \cos((B-C)/2)$.This implies $A = B-C$ or $A = C-B$.If $A = B-C$,then $A+C = B$. Since $A+B+C = 180^{\circ}$,$2B = 180^{\circ} \Rightarrow B = 90^{\circ}$.The statement in the question is $\cot \frac{A}{2} = \frac{b+c}{2}$,which is dimensionally incorrect and does not lead to $B=90^{\circ}$ as written. Thus,statement $(II)$ is false.

Observe the following statements:
$(I)$ In $\triangle ABC, b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC, \cot \frac{A}{2} = \frac{b+c}{a} \Rightarrow B = 90^{\circ}$
Which of the following is correct?

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Observe the following statements:$(I)$ In $\triangle ABC, b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$$(II)$ In $\triangle ABC, \cot \frac{A}{2} = \frac{b+c}{a} \Rightarrow B = 90^{\circ}$Which of the following is correct?