Question

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

• A. $2$
• B. $-2$
• C. $\frac{24}{7}$
• D. $\frac{7}{24}$

Answer 1

The correct option is D $\frac{7}{24}$

$\because cos\alpha isarootofthegivenequation.\therefore 25{cos}^2\alpha +5cos\alpha -12=0\Rightarrow (5cos\alpha +4)(5cos\alpha -3)=0\Rightarrow cos\alpha =-\frac{4}{5},\frac{3}{5}werejectthepositivevalue\because x<0\Rightarrow cos\alpha <0\therefore cos\alpha =-\frac{4}{5}\dots \dots \left(1\right)\Rightarrow sin\alpha =\pm \sqrt{1-co{s}^2\alpha }=\pm \sqrt{1-\frac{16}{25}}\Rightarrow sin\alpha =\pm \frac{3}{5}Werejectthepositivevalue\because x<0\Rightarrow sin\alpha <0\therefore sin\alpha =-\frac{3}{5}\dots \dots \left(2\right)From\left(1\right)and\left(2\right)tan\alpha =\frac{3}{4}Nowtan2\alpha =\frac{2tan\alpha }{1-{tan}^2\alpha }=\frac{2\times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7}\Rightarrow cot2\alpha =\frac{7}{24}Ans-OptionD$