Question

If $a$, $b$, $c$, $t$ are the solution of the equation $\tan (\theta +\frac{\pi }{4})=3tan3\theta$, no two of which have equal tangents . Then, the value of $\tan a+\tan b+\tan c+\tan t=$

• A. $1/3$
• B. $8/3$
• C. $-8/3$
• D. $0$

Answer 1

Answer: (D)

$0$

Consider the given equation,

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

Using identity ,

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\&\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

$\Rightarrow \frac{\tan \left(\frac{\pi }{4}\right)+\tan \theta }{1-\tan \left(\frac{\pi }{4}\right)\tan \theta }=3\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$

$\Rightarrow \frac{1+\tan \theta }{1-\tan \theta }=\frac{9\tan \theta -3{\tan }^3\theta }{1-3{\tan }^2\theta }$

$\Rightarrow (1+\tan \theta )(1-3{\tan }^2\theta )=(1-\tan \theta )(9\tan \theta -3{\tan }^3\theta )$

$\Rightarrow 3{\tan }^4\theta -6{\tan }^2\theta +8\tan \theta -1=0$

As $a,b,c,t,=$ are roots of this equation ,

Hence, sum of roots ,

$\tan \left(a\right)+\tan \left(b\right)+\tan \left(c\right)+\tan \left(t\right)=$ $-\frac{coeficientof{\tan }^3\theta }{coeficientof{\tan }^4\theta }$ =$-\frac{0}{1}$

$\tan \left(a\right)+\tan \left(b\right)+\tan \left(c\right)+\tan \left(t\right)=0$

Hence, this is the answer.