Q65 / 180 · COMEDK 2017 · Mathematics

The sides $a, b, c$ of a triangle are in AP. If $\cos \alpha=\frac{a}{b+c}, \cos \beta=\frac{b}{c+a}, \cos \gamma=\frac{c}{a+b}$ then $\tan ^{2} \frac{\alpha}{2}+\tan ^{2} \frac{\gamma}{2}=$

  • A
    1
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{3}$
  • Correct answer
    D
    $\frac{2}{3}$

Solution

$\frac{\cos \alpha}{1}=\frac{a}{b+c} \Rightarrow \frac{1}{\cos \alpha}=\frac{b+c}{a}$ By componendo and dividendo,we have $\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{b+c-a}{b+c+a}$ $\Rightarrow \quad \tan ^{2} \frac{\alpha}{2}=\frac{b+c-a}{3 b}\left\{\begin{array}{l}\because a, b, c \text { are in A.P. } \\ \Rightarrow 2 b=a+c\end{array}\right\}$ $\Rightarrow \quad \frac{1-\cos \gamma}{1+\cos \gamma}=\frac{a+b-c}{a+b-c}$ $\Rightarrow \quad \tan ^{2} \frac{\gamma}{2}=\frac{a+b-c}{3 b}$ $\therefore \tan ^{2} \frac{\alpha}{2}+\tan ^{2} \frac{\gamma}{2}=\frac{2 b}{3 b}=\frac{2}{3}$

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

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