Trigonometric functions of sine, cosine, tangent, cotangent are based on the signs of y and x coordinates in the respective four quadrants.

θ lies in which Quadrant |
I |
II |
III |
IV |

Trigonometric functions |
||||

Sin θ |
+ve |
+ve |
-ve |
-ve |

Cos θ |
+ve |
-ve |
-ve |
+ve |

Tan θ |
+ve |
-ve |
+ve |
-ve |

Cot θ |
+ve |
-ve |
+ve |
-ve |

Cosec θ |
+ve |
+ve |
-ve |
-ve |

Sec θ |
+ve |
-ve |
-ve |
+ve |

From the above,

**Quadrant I: **The values of the trigonometric functions including Sine, Cosine, Tangent, Cotangent of any of the arc from the I Quadrant are the positive ones as coordinates are positive of the given points – P, S₁, and S₂, which define their specific values.

**Quadrant II: **From the 2nd quadrant points, for the arcs, P & S₂, both consist of negative abscissas (in the figure above), so, cotangent and the cosine exist as negative. Terminal point P’s ordinate is +ve so that Sine is +ve while ordinate of Point S₁ is -ve and hence, the tangent is negative.

** (**

**Quadrant III: **As abscissas plus terminal points P’s ordinates of arcs from the 3rd quadrant (see the figure above) are -ve. It follows the sine and cosine functions of the given arcs are -ve. Point S1’s ordinates plus point S2’s abscissa which belongs to arc from the III quadrant are +ve. Therefore, cotangent and tangent of the given arcs are +ve.

**Quadrant IV: **The functions including tangent, sine, and cotangent of the given arcs from IVth quadrant are -ve, similar to points S₁, S₂, and P’s coordinates which do belongs to them only. Only arc’s cosine functions from IVth quadrant is +ve are point P’s abscissas which belongs to them as you can see in the figure above.

Quadrant |
Values of arc |
Sin x |
Cos x |
Tan x |
Cot x |

I |
From 0 to 90º |
+ |
+ |
+ |
+ |

II |
From 90º to 180º |
+ |
– |
– |
– |

III |
From 180º to 270º |
– |
– |
+ |
+ |

IV |
From 270º to 360º |
– |
+ |
– |
– |

**Example**

**Question 1: Give signs of the trigonometric functions**

Solution: Sine 146º = Sin (90º + 56º) = As Sine lies from 0º to 90º, therefore, The sign will be positive.

**Question 2: Cos 455º**

Solution: Cos (360**º **+ 95º) = Cos (90º + 5º) = Cos 5º

From the above solution, it clearly means that the sign is positive.

**Question 3: Tan 573º**

Solution: Tan (360**º** + 213**º**) = Tan (180**º** + 33**º**) = Tan 33**º**

Hence, Tan 573º is positive