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

2 Answers
Dec 14, 2016

Answer:

#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]}

Dec 14, 2016

Answer:

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