Question

The value of $\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }+\frac{1}{\tan \gamma }+\frac{1}{\tan \delta }$ is

• A. $-8$
• B. $8$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $8$

$\tan (\theta +\frac{\pi }{4})=3\tan 3\theta$

$\Rightarrow \frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=3\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$\Rightarrow {\tan }^4\theta -2{\tan }^2\theta +\frac{8}{3}\tan \theta -\frac{1}{3}=0$

$\Rightarrow {\tan }^4\theta +0{\tan }^3\theta -2{\tan }^2\theta +\frac{8}{3}-\frac{1}{3}=0$

Sum roots taken $3$ at time is

$\tan \beta \cdot \tan \gamma \cdot \tan \delta +\tan \alpha \cdot \tan \gamma \cdot \tan \delta +\tan \alpha \cdot \tan \beta \cdot \tan \delta +\tan \alpha \cdot \tan \beta \cdot \tan \gamma =-\frac{8}{3}$

Product of roots is

$\tan \alpha \cdot \tan \beta \cdot \tan \gamma \cdot \tan \delta =-\frac{1}{3}$

Therefore

$\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }+\frac{1}{\tan \gamma }+\frac{1}{\tan \delta }$

$=\frac{\tan \beta \cdot \tan \gamma \cdot \tan \delta +\tan \alpha \cdot \tan \gamma \cdot \tan \delta +\tan \alpha \cdot \tan \beta \cdot \tan \delta +\tan \alpha \cdot \tan \beta \cdot \tan \gamma }{\tan \alpha \cdot \tan \beta \cdot \tan \gamma \cdot \tan \delta }$

$=\frac{-\frac{8}{3}}{-\frac{1}{3}}=+8$