How do you find the value of cot(theta/2) given tantheta=-7/24 and (3pi)/2<theta<2pi?

Jun 22, 2016

$\cot \left(\frac{\theta}{2}\right) = - 7.$

Explanation:

We know that,
$\tan \theta = \frac{2 \tan \left(\frac{\theta}{2}\right)}{1 - {\tan}^{2} \left(\frac{\theta}{2}\right)}$

So, letting $\tan \left(\frac{\theta}{2}\right) = t$, & using given value of $\tan \theta = - \frac{7}{24}$ in the LHS, we get the eqn.,

$- \frac{7}{24} = \frac{2 t}{1 - {t}^{2}} ,$ or,$- 7 + 7 {t}^{2} = 48 t ,$ i.e., $7 {t}^{2} - 48 t - 7 = 0.$
$\therefore \left(t - 7\right) \left(7 t + 1\right) = 0 , .$ which gives $t = \tan \left(\frac{\theta}{2}\right) = 7 , \mathmr{and} , - \frac{1}{7.}$

Clearly, $\cot \left(\frac{\theta}{2}\right) = \frac{1}{7} , \mathmr{and} , - 7.$

Given that $3 \frac{\pi}{2}$<$\theta$<$2 \pi .$ $\Rightarrow$ $3 \frac{\pi}{4}$<$\left(\frac{\theta}{2}\right)$<$\pi ,$ meaning, $\frac{\theta}{2}$ lies in the IInd Quadrant, where cot is $- v e .$

Finally, $\cot \left(\frac{\theta}{2}\right) \ne \frac{1}{7} , b u t , = - 7.$