Question

Suppose tan(theta) = 4/3 and csc<0, find tan(theta/2)

Answer 1

Using half-angle formula.

$\tan\theta=\frac{2t}{1-t^2}$ , where $t=\tan\frac{\theta}{2}$ .

$\frac{4}{3}=\frac{2t}{1-t^2}\\ 4(1-t^2)=6t\\ 2(1-t^2)=3t\\ 2-2t^2=3t\\ 2t^2+3t-2=0\\ 2t^2+4t-t-2=0\\ 2t(t+2)-1(t+2)=0\\ (2t-1)(t+2)=0\\ 2t-1=0 ,t+2=0\\ t=\frac{1}{2}, t=-2$

So,

$\tan\frac{\theta}{2}=\frac{1}{2} or-2$

But,. there is a condition that $\csc\theta<0$ .

If $\tan\frac{\theta}{2}=\frac{1}{2}$ , then $\csc\theta=\frac{5}{4}>0$ . Therefore, this is not the answer.

If $\tan\frac{\theta}{2}=-2,$ then $\csc\theta=-\frac{5}{4}<0$ . Therefore this is the answer.

So, $\tan\frac{\theta}{2}=-2.$