# Thread: Exact values in radians of θ, goven cosec & tan

1. ## Exact values in radians of θ, goven cosec & tan

Given cosecθ= -2, tanθ= 1/3√3 and -π < θ < π.
I must find the exact value of θ in radians, justifying my answer.

It shouldn't be too difficult;
cosec theta = -2 implies sin theta = -(1/2); value of tan theta cannot be (1)/(3)√(3) but if it is taken as 1/√(3); then required theta = -(5*pi/6).

Or is this incorrect?

2. ## Re: Exact values in radians of θ, goven cosec & tan

this appears to be in Q3 not sure what you mean by -π < θ < π.

both of these are $\displaystyle \frac{-5\pi}{6}$ in Q3

3. ## Re: Exact values in radians of θ, goven cosec & tan

Hello, froodles01!

$\displaystyle \csc\theta = \text{-}2,\;\tan\theta= \tfrac{\sqrt{3}}{3}\,\text{and }\,\text{-}\pi \,<\,\theta\,<\,\pi$

$\displaystyle \text{Find the exact value of }\theta\text{ in radians.}$

$\displaystyle \csc\theta \,=\,\text{-}2 \quad\Rightarrow\quad \sin\theta \,=\,\text{-}\tfrac{1}{2} \quad\Rightarrow\quad \theta \:=\:\text{-}\tfrac{5\pi}{6},\;\text{-}\tfrac{11\pi}{6}$

Since $\displaystyle \tan\theta$ is positive, $\displaystyle \theta$ is in Quadrant 1 or 3.

Therefore: .$\displaystyle \theta \:=\:-\frac{5\pi}{6}$