# If costheta+csctheta>0, in which quadrant does theta lie?

Dec 14, 2016

$\theta$ lies in first and second quadrant.

General solution is $\left(2 n + 1\right) \pi < \theta < 2 n \pi$, where $n$ is an integer

#### Explanation:

$\cos \theta + \csc \theta > 0$

$\Leftrightarrow \cos \theta + \frac{1}{\sin} \theta > 0$

or $\frac{\sin \theta \cos \theta + 1}{\sin} \theta > 0$

or $\frac{2 \sin \theta \cos \theta + 2}{2 \sin \theta} > 0$

or $\frac{\sin 2 \theta + 2}{2 \sin \theta} > 0$

As $- 1 \le \sin 2 \theta \le + 1$,

numerator $\sin 2 \theta + 2$ is always positive

Hence, for $\cos \theta + \csc \theta > 0$,

we should have $2 \sin \theta > 0$ or $\sin \theta > 0$

and hence $\theta$ lies in first and second quadrant.

General solution is $\left(2 n + 1\right) \pi < \theta < 2 n \pi$, where $n$ is an integer
graph{cosx+cscx [-10.42, 9.58, -1.84, 8.16]}

Dec 14, 2016

#### Explanation:

$f \left(t\right) = \cos t + \frac{1}{\sin} t$
Determine the sign of f(t) by finding the sign of cos t and sin t in each quadrant.
Quadrant I --> cos t > 0 and sin t > 0
There for f(t) > 0
Quadrant II --> cos t < 0 and sin t > 0.
There for f(t) < 0
Quadrant III --> cos t < 0 and sin t < 0.
Therefor, f(t) < 0
Quadrant IV --> cos t > 0 and sin t < 0.
There for, f(t) < 0.