Question

Let $\alpha$ be the root of the equation $25{\cos }^2\theta +5\cos \theta -12=C$, where $\frac{\pi }{2}<\alpha <\pi$.

What is $\tan \alpha$ equal to?

• A. $-\frac{3}{4}$
• B. $\frac{3}{4}$
• C. $-\frac{4}{3}$
• D. $-\frac{4}{5}$

Answer 1

The correct option is A $-\frac{3}{4}$

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

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

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

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

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

Since $\alpha$ is the root of the equation and it lies between $\frac{\pi }{2}<\alpha <\pi \Rightarrow \cos \alpha$ is '-ve'.

Thus $\alpha ={\cos }^{-1}\left(\frac{-4}{5}\right)$

Since it lies in the range $\frac{\pi }{2}<\alpha <\pi$, so $\tan \alpha$ is '-ve'

$\tan \alpha =\tan \left[{\cos }^{-1}\left(\frac{-4}{5}\right)\right]$

$=\tan \left[{\tan }^{-1}(-\frac{3}{4})\right]$

$=-\frac{3}{4}$