Q88 / 200 · Manipal (MET) 2010 · Mathematics

In a $\triangle A B C, \angle B=90^{\circ}$, then $\tan ^2\left(\frac{A}{2}\right)$ is

  • Correct answer
    A
    $\frac{b-c}{b+c}$
  • B
    $\frac{b+c}{b-c}$
  • C
    $\frac{b-2 c}{b+c}$
  • D
    None of these

Solution

In given right angled triangle, we have $\cos A=c / b$ $\Rightarrow \frac{1-\tan ^2 \frac{A}{2}}{1+\tan ^2 \frac{A}{2}}=c / b$ $\Rightarrow \frac{\left(1+\tan ^2 \frac{A}{2}\right)-\left(1-\tan ^2 \frac{A}{2}\right)}{\left(1+\tan ^2 \frac{A}{2}\right)+\left(1-\tan ^2 \frac{A}{2}\right)}=\frac{(b-c)}{(b+c)}$ (by componendo and dividendo rule) Hence, $\tan ^2 \frac{A}{2}=\frac{b-c}{b+c}$

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

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