Q103 / 160 · WBJEE 2009 · Mathematics

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

  • A
    $\frac{2 a}{b}$
  • Correct answer
    B
    $\frac{2 b}{a}$
  • C
    $\frac{a}{b}$
  • D
    $\frac{b}{a}$

Solution

Hints : Let $\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)=\theta$, then $\cos 2 \theta=\frac{a}{b}$ $$ \begin{aligned} & \tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right] \\ & =\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)=2\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)=\frac{2}{\cos 2 \theta}=\frac{2}{a / b}=\frac{2 b}{a} \end{aligned} $$

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

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