when is tanx+cotx<or=1

Question

The question is: When is tanx + cot x ≤ 1

Answer 1

There are some trig calculators on line, where you can plug in this problem. I tried one and got a "can't solve" message back. That suggests something is unusual about this problem

Let y = tanx = 1/cotx

tanx + cotx < 1

y+ 1/y < 1

multiply through by y and solve for y

y² +1 < y

y²-y+1 < 0 has no real solutions. Try applying the quadratic formula and the discriminate is negative.

That's true for the 1st and 3rd quadrants. You can almost see this intuitively. Start with 45 degrees, where tanx = cot x = 1 They sum to 2 > 1. If you make the angle any larger or smaller either tanx or cotx will be > 1.

That leaves solutions in the 2nd and 4th quadrants, where tanx and cotx are negative. Add two negatives, they're always negative, and less than 1

Answer is 90<x<180 and 270<x<360

or any integer multiple of 360 added to those numbers

Answer 2

SInce tan(x) and cot(x) are always positive (or at least zero) in the first and third quadrants, or, from [0,Π/2] and [Π,3Π/2], then their sum will be greater than 1 in those intervals. Similarly, their sum will be less than or equal to 1 in quadrants ll and lV, or from [π/2,π]±πk.

We have to understand that tan(x) and cot(x) are reciprocal functions, so when added together, even if one of them approaches zero, the other one will approach infinity.

If you graph the sum of tan(x) and cot(x), you will see that this pattern repeats itself for ±Πk

Long story short, x=[π/2/,π] ±πk, k∈Ζ