Question

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

Answer 1

`theta` lies in first and second quadrant.

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

Explanation:

`costheta+csctheta>0`

`hArrcostheta+1/sintheta>0`

or `(sinthetacostheta+1)/sintheta>0`

or `(2sinthetacostheta+2)/(2sintheta)>0`

or `(sin2theta+2)/(2sintheta)>0`

As `-1<=sin2theta<=+1`,

numerator `sin2theta+2` is always positive

Hence, for `costheta+csctheta>0`,

we should have `2sintheta>0` or `sintheta>0`

and hence `theta` lies in first and second quadrant.

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

Answer 2

Quadrant I

Explanation:

`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