If $cos θ + csc θ > 0$ $costheta+csctheta>0$ , in which quadrant does $θ$ $theta$ lie?

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If $cos θ + csc θ > 0$ $costheta+csctheta>0$ , in which quadrant does $θ$ $theta$ lie?

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Answer 1 · 2016-12-14T14:08:18

$θ$ $theta$ lies in first and second quadrant.

General solution is $(2 n + 1) π < θ < 2 n π$ $(2n+1)pi< theta< 2npi$ , where $n$ $n$ is an integer

Explanation:

$cos θ + csc θ > 0$ $costheta+csctheta>0$

$\Leftrightarrow cos θ + \frac{1}{sin θ} > 0$ $hArrcostheta+1/sintheta>0$

or $\frac{sin θ cos θ + 1}{sin θ} > 0$ $(sinthetacostheta+1)/sintheta>0$

or $\frac{2 sin θ cos θ + 2}{2 sin θ} > 0$ $(2sinthetacostheta+2)/(2sintheta)>0$

or $\frac{sin 2 θ + 2}{2 sin θ} > 0$ $(sin2theta+2)/(2sintheta)>0$

As $- 1 \leq sin 2 θ \leq + 1$ $-1<=sin2theta<=+1$ ,

numerator $sin 2 θ + 2$ $sin2theta+2$ is always positive

Hence, for $cos θ + csc θ > 0$ $costheta+csctheta>0$ ,

we should have $2 sin θ > 0$ $2sintheta>0$ or $sin θ > 0$ $sintheta>0$

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

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

Answer 2 · 2016-12-14T14:47:14

Quadrant I

Explanation:

$f (t) = cos t + \frac{1}{sin t}$ $f(t) = cos t + 1/sin t$
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.
Answer: f(t) > 0 in Quadrant I

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If $cos θ + csc θ > 0$ $costheta+csctheta>0$ , in which quadrant does $θ$ $theta$ lie?

2 Answers

Explanation:

Explanation:

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Science

Math

Humanities

... and beyond

If cosθ+cscθ>0costheta+csctheta>0, in which quadrant does θtheta lie?

2 Answers

Explanation:

Explanation:

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If $cos θ + csc θ > 0$ $costheta+csctheta>0$ , in which quadrant does $θ$ $theta$ lie?