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(s-c) + s(s-b)}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For statement $(I)$: $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = b \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-b)}{ac} = \frac{s(s-c) + s(s-b)}{a} = \frac{s(2s - b - c)}{a} = \frac{s(a)}{a} = s$. Thus,statement $(I)$ is true.For statement $(II)$: Given $\cot \frac{A}{2} = \frac{b+c}{a}$. Using the sine rule,$\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A} = \frac{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}}$. Since $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$,$\sin \frac{B+C}{2} = \cos \frac{A}{2}$. So,$\cot \frac{A}{2} = \frac{\cos \frac{A}{2} \cos \frac{B-C}{2}}{\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\cos \frac{B-C}{2}}{\sin \frac{A}{2}}$. This implies $\cos \frac{A}{2} = \cos \frac{B-C}{2}$,so $\frac{A}{2} = \frac{B-C}{2} \implies A = B - C \implies A+C = B$. Since $A+B+C = 180^{\circ}$,$2B = 180^{\circ} \implies B = 90^{\circ}$. The statement in the question says $\cot \frac{A}{2} = \frac{b+c}{2}$,which is dimensionally incorrect as the denominator should be $a$. 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 $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
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 $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$Which of the following is correct?