Question

The solution set of the equation $\tan (\pi \: \tan\: x) = \cot (\pi \cot \: x) $ is

• A. $ \phi$
• B. {0}
• C. $\left\{\frac{\pi}{4}\right\}$
• D. none of these.

Answer 1

Answer: (A)

$\tan \, (\pi \, \tan\, x) = \cot (\pi\, \cot\, x) = \tan \left( \frac{\pi}{2} = \pi \, \cot \, x\right)$
$\Rightarrow $ $\pi\, \tan\, x = \frac{\pi}{2} - \pi \, \cot\, x$
$\Rightarrow $ $ \tan \,x + \cot\, x = \frac{1}{2}$
$\Rightarrow $ $ \tan \,x + \cot\, x = \frac{1}{2 } \tan\, x +\frac{1}{\tan\, x} = \frac{1}{2} \Rightarrow \, 2 \tan^{2} x + 2 = \tan \,x$
$\Rightarrow $ $ 2\, \tan^{2 }x - \tan \,x + 2 = 0$
This is quadratic in $\tan\, x$
$\Rightarrow $ $\tan \, x = \frac{ 1 \pm \sqrt{1 - 4.4 }}{4} = \frac{ 1 \pm \sqrt{-15}}{2}$
which are not real solution set = $\phi$