Question

If $\cos \alpha$ is a root of $25x^2+5x-12=0$ and $(-1<x<0)$, then the value of $\tan 2\alpha$ can be:

• A. $\frac{24}{12}$
• B. $-\frac{12}{24}$
• C. $\frac{20}{24}$
• D. $\frac{24}{7}$

Answer 1

Answer: (D)

$\frac{24}{7}$

$\cos \alpha$ is a root of the given equation.
$\Rightarrow 25{\cos }^2\alpha +5\cos \alpha -12=0\Rightarrow (5\cos \alpha +4)(5\cos \alpha -3)=0\Rightarrow \cos \alpha =[-\frac{4}{5},\frac{3}{5}]\Rightarrow \tan \alpha =\frac{\pm \sqrt{1-{\cos }^2\alpha }}{\cos \alpha }=\frac{\pm \sqrt{1-\frac{16}{25}}}{-\frac{4}{5}}=\pm \frac{3}{4}$
We reject the negative value as $\tan \alpha$ is positive within the given range of $x.$
Now,
$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{2\times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7}$
Ans: Option D.