Q72 / 160 · TS EAMCET 2008 · Mathematics

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?

  • A
    Both I and II are true.
  • Correct answer
    B
    I is true, II is false.
  • C
    I is false, II is true.
  • D
    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.

Hence, the correct answer is option: 'B'.

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