__Trigonometric ratios of supplementary angles :__

Two angles are supplementary to each other if their sum is equal to 180°.

Trigonometric-ratios of supplementary angles are given below.

sin (180° - θ) = sin θ cos (180° - θ) = - cos θ tan (180° - θ) = - tan θ csc (180° - θ) = csc θ sec (180° - θ) = - sec θ cot (180° - θ) = - cot θ |
sin (180° + θ) = - sin θ cos (180° + θ) = - cos θ tan (180° + θ) = tan θ csc (180° + θ) = - csc θ sec (180° + θ) = - sec θ cot (180° + θ) = cot θ |

Let us see, how the trigonometric ratios of supplementary angles are determined.

To know that, first we have to understand ASTC formula.

The ASTC formula can be remembered easily using the following phrases.

**"All Sliver Tea Cups" **

or

**"All Students Take Calculus"**

ASTC formla has been explained clearly in the figure given below.

More clearly

From the above picture, it is very clear that

**(i) (180° - θ) falls in the second quadrant and **

**(i) (180° + θ) falls in the third quadrant**

In the second quadrant (180° - θ), sin and csc are positive and other trigonometric ratios are negative.

In the third quadrant (180° + θ), tan and cot are positive and other trigonometric ratios are negative.

When we have the angles 90° and 270° in the trigonometric ratios in the form of

(90° + θ)

(90° - θ)

(270° + θ)

(270° - θ)

We have to do the following conversions,

sin θ <------> cos θ

tan θ <------> cot θ

csc θ <------> sec θ

For example,

sin (270° + θ) = - cos θ

cos (90° - θ) = sin θ

For the angles 0° or 360° and 180°, we should not make the above conversions.

**Problem 1 :**

Evaluate : sin (180° - θ)

**Solution :**

To evaluate sin (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "sin" will not be changed as "cos"

(iii) In the II nd quadrant, the sign of "sin" is positive.

Considering the above points, we have

sin (180° - θ) = sin θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 2 :**

Evaluate : cos (180° - θ)

**Solution :**

To evaluate cos (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "cos" will not be changed as "sin"

(iii) In the II nd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° - θ) = - cos θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 3 :**

Evaluate : tan (180° - θ)

**Solution :**

To evaluate tan (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "tan" will not be changed as "cot"

(iii) In the II nd quadrant, the sign of "tan" is negative.

Considering the above points, we have

tan (180° - θ) = - tan θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 4 :**

Evaluate : csc (180° - θ)

**Solution :**

To evaluate csc (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "csc" will not be changed as "sec"

(iii) In the II nd quadrant, the sign of "csc" is positive.

Considering the above points, we have

csc (180° - θ) = csc θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 5 :**

Evaluate : sec (180° - θ)

**Solution :**

To evaluate sec (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "sec" will not be changed as "csc"

(iii) In the II nd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° - θ) = - sec θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 6 :**

Evaluate : cot (180° - θ)

**Solution :**

To evaluate cot (180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "cot" will not be changed as "tan"

(iii) In the II nd quadrant, the sign of "cot" is negative.

Considering the above points, we have

cot (180° - θ) = - cot θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 7 :**

Evaluate : sin (180° + θ)

**Solution :**

To evaluate sin (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "sin" will not be changed as "cos"

(iii) In the III rd quadrant, the sign of "sin" is negative.

Considering the above points, we have

sin (180° + θ) = - sin θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 8 :**

Evaluate : cos (180° + θ)

**Solution :**

To evaluate cos (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "cos" will not be changed as "sin"

(iii) In the III rd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° + θ) = - cos θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 9 :**

Evaluate : tan (180° + θ)

**Solution :**

To evaluate tan (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "tan" will not be changed as "cot"

(iii) In the III rd quadrant, the sign of "tan" is positive.

Considering the above points, we have

tan (180° + θ) = tan θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 10 :**

Evaluate : csc (180° + θ)

**Solution :**

To evaluate csc (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "csc" will not be changed as "sec"

(iii) In the III rd quadrant, the sign of "csc" is negative.

Considering the above points, we have

csc (180° + θ) = - csc θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 11 :**

Evaluate : sec (180° + θ)

**Solution :**

To evaluate sec (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "sec" will not be changed as "csc"

(iii) In the III rd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° + θ) = - sec θ

Let us look at the next stuff on "Trigonometric ratios of supplementary angles"

**Problem 12 :**

Evaluate : cot (180° + θ)

**Solution :**

To evaluate cot (180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "cot" will not be changed as "tan"

(iii) In the III rd quadrant, the sign of "cot" is positive.

Considering the above points, we have

cot (180° + θ) = cot θ

sin (180° - θ) = sin θ cos (180° - θ) = - cos θ tan (180° - θ) = - tan θ csc (180° - θ) = csc θ sec (180° - θ) = - sec θ cot (180° - θ) = - cot θ |
sin (180° + θ) = - sin θ cos (180° + θ) = - cos θ tan (180° + θ) = tan θ csc (180° + θ) = - csc θ sec (180° + θ) = - sec θ cot (180° + θ) = cot θ |

**Click here to know more about ASTC formula**

After having gone through the stuff given above, we hope that the students would have understood "Trigonometric ratios of supplementary angles"

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