Question

If $\alpha$ is a root of $25{\cos }^2\theta +5\cos \theta =0,\frac{\pi }{2}<\alpha <\pi$, then ${\sin }^2\alpha$ is equal to-

• A. $\frac{24}{25}$
• B. $-\frac{24}{25}$
• C. $\frac{13}{18}$
• D. $-\frac{13}{18}$

Answer 1

The correct option is A $\frac{24}{25}$

$25{\cos }^2\theta +5\cos \theta -12=0$

$\Rightarrow 25{\cos }^2\theta +20\cos \theta -15\cos \theta -12=0$

$\Rightarrow (5\cos \theta -3)(5\cos \theta +4)=0$

$\Rightarrow \cos \theta =\frac{3}{4},-\frac{4}{5}$

$\because x$ lies between $\frac{\pi }{2}$ and $\pi$

The root will be negative

$\therefore \cos \alpha =\frac{-4}{5}$

${\sin }^{2}x=1-{\cos }^2\alpha$ $[\because {\sin }^2\theta +{\cos }^2\theta =1]$

$=1-{\left(\frac{-4}{5}\right)}^2$

$=1-\frac{16}{25}$

$=\frac{9}{25}$