Question

If $\cos \theta$ and $\cos 2\theta$ are the roots of equation $27x^2-15x-2=0,$ then $\tan 3\theta$ is equal to :-

• A. $\frac{7\sqrt{5}}{22}$
• B. $-\frac{7\sqrt{5}}{22}$
• C. $\frac{7\sqrt{5}}{2}$
• D. $-\frac{7\sqrt{5}}{2}$

Answer 1

Answer: (B)

$-\frac{7\sqrt{5}}{22}$

${27x}^2-15x-2=0$

Sum of roots $=\cos \theta +\cos 2\theta =\frac{15}{27}$

$\cos \theta +{2\cos }^2\theta -1=\frac{15}{9}$

${2\cos }^2\theta +\cos \theta -1-\frac{5}{9}=0$

${2\cos }^2\theta +\cos \theta -\frac{14}{9}=0$

$18{\cos }^2\theta +\cos \theta -\frac{14}{9}=0$

${18\cos }^2\theta +9\cos \theta -14=0$

$\cos \theta =\frac{-9\pm \sqrt{81+4\left(14\right)\left(18\right)}}{2\times 18}$

$=\frac{-9\pm 33}{2\times 18}=\frac{24}{36},\frac{-42}{36}$

$\cos \theta =\frac{2}{3},\frac{-7}{6}$

$\times$ not possible

$\because$ $-1\le \cos \theta \subseteq 1$

$\therefore$, $cos\theta =2/3$

In $\Delta ABC\cos \theta =2/3$

${AC}^2\ne {AB}^2+{BC}^2$

${AB}^2=5$

$AB=\sqrt{5}\Rightarrow \tan \theta =\frac{\sqrt{5}}{2}$

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

$=\frac{3\left(\frac{\sqrt{5}}{2}\right)-\frac{5\sqrt{5}}{8}}{1-3\left(\frac{5}{4}\right)}=\frac{\frac{3\sqrt{5}}{2}-\frac{5\sqrt{5}}{8}}{1-\frac{15}{4}}$

$=\frac{7\sqrt{5}}{8}\times \frac{4}{11}=\frac{-7\sqrt{5}}{22}$