Question

If $\tan a,\tan b,\tan c,\tan d$ are the roots of the eqn $\tan (45°+\theta )=3\tan 3\theta$, then
$\frac{1}{\tan a}+\frac{1}{\tan b}+\frac{1}{\tan c}+\frac{1}{\tan d}$ is

Answer 1

$\tan (45°+\theta )=3\tan 3\theta \Rightarrow \frac{1+\tan \theta }{1-\tan \theta }=3\cdot \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$
Let $\tan \theta =t$
then,
$\frac{1+t}{1-t}=3\cdot \frac{3t-t^3}{1-3t^2}\Rightarrow (1+t)(1-3t^2)=3(3t-t^3)(1-t)\Rightarrow 1-3t^2+t-3t^3=9t-9t^2-3t^3+3t^4\Rightarrow 3t^4-6t^2+8t-1=0$
$\because \tan a,\tan b,\tan c,\tan d$ are roots of the above equation
$\therefore$ Equation whose roots are $\frac{1}{\tan a},\frac{1}{\tan b},\frac{1}{\tan c},\frac{1}{\tan d}$ is
$-t^4+8t^3-6t^2+3=0\Rightarrow t^4-8t^3+6t^2-3=0$

Hence, the sum of the roots
$\frac{1}{\tan a}+\frac{1}{\tan b}+\frac{1}{\tan c}+\frac{1}{\tan d}=8$