Sign, Domain and range of T-ratios
Sign of t-ratios
Quadrant | I | II | III | IV |
T-ratios which are +ve | All |
sin cosec |
tan cot |
cos sec |
This table can be memorised with the help of phrase "Add Sugar To Coffee"
Add | Sugar | To | Coffee |
All | Sin | Tan | Cos |
I | II | III | IV |
Domain of T-ratios
Function | Domain |
sin, cos | all real numbers |
tan, sec | all real numbers other than (2 n +1) /2, n Z |
cot, cosec | all real numbers other than n , n Z |
Limits of T-ratios (i.e. range)
The maximum and minimum values of sin and cos
are +1 and -1 respectively.
Thus, -1 cos
1 and -1 sin 1
Function | Range |
sin, cos | [-1, 1] |
tan, cot | any real value |
sec, cosec | any real value except (-1, 1) |
Illustrative Examples
Example
If tan = -2, find the values of the remaining trigonometric ratios of .
Solution
Given tan = -2 which is - ve, therefore,
lies in second or fourth quadrant.
Also sec² = 1 + tan² = 1 +(-2)² =
5
=> sec = ±5
Two cases arise:
Case I. When lies in the second quadrant, sec
is - ve.
sec = -5 =>
cos = -1/5
sin = (sin/cos).cos
= tan. cos = (-2).(-1/5)
= 2/5
=> cosec = (5)/2
Also tan = -2 => cot
= -1/2
Case II. When lies in the fourth quadrant, sec
is + ve.
sec = 5 =>
cos = 1/5
sin = (sin/cos).cos
= tan. cos = (-2).(1/5)
= -2/5
=> cosec = -(5)/2
Also tan = -2 => cot
= -1/2
Example
Prove that sin = x + 1/x is not possible for real x.
Solution
When x > 0, x +1/x = (x - 1/x)² +2
2
When x < 0, let x = -y where y > 0. Then
Thus x +1/x 2 or x +1/x
-2
=> sin 2
or sin -2
Also, we know that -1 sin
1
Hence sin = x +1/x is not possible for any real x.
Aliter.
sin = x +1/x => x 2
-sin . x +1 = 0
It is a quadratic in x. As x is real, it has real roots
=> (-sin )² -4. 1. 1
0
(discriminant 0)
=> sin² 4 =>
|sin | 2
=> sin -2 or sin
2
Also we know that 1 sin
-1
Hence sin = x +1/x is not possible for any real x.
Example
Is the equation 2 sin² -cos +4 = 0 possible?
Solution
2 sin² -cos +4 = 0
=> 2 (1 -cos² ) -cos
+4 = 0
=> -2 cos² +cos -6 = 0
=> 2 cos² +cos -6 = 0
=> (2 cos -3)(cos + 2) = 0
=> 2 cos -3 = 0 or cos +2 = 0
=> cos = 3/2 or cos
= -2, both of which are impossible as -1 cos
1.
Hence the equation 2 sin² -cos +4
= 0 is not possible.
Exercise
- Which of the six t-ratios are positive for the angles
(i) 240° (ii) -420°? - Find the other five t-ratios if
(i) cos A = 1/2 and A lies in the second quadrant
(ii) sin A = 3/5 and /2 < A <
(iii) tan A = 3/2 and A does not lie in first quadrant
(iv) cot A = 12/5 and < A < 3/2 - In which quadrant does lie if
(i) cos is positive and tan is negative
(ii) both sin and cos are negative
(iii) sin = 4/5 and cos = 3/5
(iv) sin = 2/3 and cos = 1/3 - For what real values of x is the equation 2 cos = x +1/x possible?
- If sin sec = -1 and lies in the second quadrant, find sin and sec .
- If sec A = x +1/4x, prove that sec A +tan A = 2 x or 1/2x.
- If sin = 12/13 and lies in the
second quadrant, show that
sec +tan = -5. - If sin : cos : : 3 : 1, find sin , cos .
Answers
1. (i) tan, cot (ii) cos, sec2. (i) sin A = 3/2 , tan A = -3, cot A = -1/3, sec A = -2, cosec A = 2/3
(ii) cos A = 4/5, tan A = 3/4, cot A = 4/3, sec A = 5/4, cosec A = 5/3
(iii) sin A = 3/5, cos A = 4/5, cot A = 4/3, sec A = 5/4 , cosec A = 5/3
(iv) sin A = 5/13 , cos A = 12/13, tan A = 5/12 , sec A = 13/12, cosec A = -13/5
3. (i) fourth (ii) third (iii) second
(iv) not possible as we must have sin² +cos² = 1
4. x = ±1 only
5. 1/2, -2
8. sin = 3/2, cos = 1/2 or sin = - 3/2, cos = 1/2