| 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 |
| Function | Domain |
| sin, cos | all real numbers |
| tan, sec | all real numbers other than
(2 n +1) |
| cot, cosec | all real numbers other than
n |
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) |
If tan
= -2, find the values of the
remaining trigonometric ratios of
.
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
Prove that sin
= x + 1/x is not
possible for real x.
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.
Is the equation 2 sin²
-cos
+4 = 0 possible?
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.