__maths__>__Trigonometry__>__Trigonometric Identities and Complementary Angles__### Trigonometric Ratios for Complementary Angles

The relationship between trigonometric ratios of an angle and its complementary angle is explained.

*click on the content to continue..*

What is the complementary angle of `theta`?

- angle that completes `theta` to a right angle
- `90-theta`
- both the above
- both the above

The answer is 'both the above'

In the right angled triangle shown in figure, if `/_P = theta` then what is `/_R` ?

- complementary angle of `theta`
- `90-theta`
- both the above
- both the above

The answer is 'both the above'

In the right angled triangle shown in figure, what is the opposite side for `/_P ` ?

- `PQ`
- `QR`
- `QR`

The answer is '`QR`'

In the right angled triangle shown in figure, what is the adjacent side for `/_R ` ?

- `PQ`
- `QR`
- `QR`

The answer is 'QR'

A right angled triangle is shown in figure. Opposite of `/_theta` = Adjacent of `/_(90-theta)`

Adjacent of `/_theta` = Opposite of `/_(90-theta)`

The trigonometric ratios of `90-theta` can be expressed in terms of trigonometric ratios of `theta` or vice versa. From the figure

`sin theta = (RQ)/(PR)`

`cos theta = (PQ)/(PR)`

`sin (90-theta) = (PQ)/(PR)`

`cos (90-theta) = (RQ)/(PR)`

Comparing the above, it is concluded that

`sin (90-theta) = cos theta`

`cos (90-theta) = sin theta`

It is noted that the two angles in a right angles triangle are complementary angles and the trigonometric ratios of one angle are related to the trigonometric ratios of the other angle.

That is

• hypotenuse is common for both the angles

• opposite side for an angle is the adjacent side for the other angle

• adjacent side for an angle is the opposite side for the other angle

This relation is used to derived **Trigonometric Ratios for Complementary Angles **.

**Trigonometric Ratios for Complementary Angles: **

`sin(90-theta) = cos theta`

`cos(90-theta) = sin theta`

The trigonometric ratios of `90-theta` can be expressed in terms of trigonometric ratios of `theta` or vice versa. From the figure

`tan theta = (RQ)/(PQ)`

`cot theta = (PQ)/(RQ)`

`tan (90-theta) = (PQ)/(RQ)`

`cot (90-theta) = (RQ)/(PQ)`

Comparing the above, it is concluded that

`tan (90-theta) = cot theta`

`cot (90-theta) = tan theta`

`tan(90-theta) = cot theta`

`cot(90-theta) = tan theta`

The trigonometric ratios of `90-theta` can be expressed in terms of trigonometric ratios of `theta` or vice versa. From the figure

`csc theta = (PR)/(RQ)`

`sec theta = (PR)/(PQ)`

`csc (90-theta) = (PR)/(PQ)`

`sec (90-theta) = (PR)/(RQ)`

Comparing the above, it is concluded that

`csc(90-theta) = sec theta`

`sec(90-theta) = csc theta`

`csc(90-theta) = sec theta`

`sec(90-theta) = csc theta`

*Solved Exercise Problem: *

Which of the following equals `sin 36`?

- `cos 54`
- `cos 54`
- `tan 54`
- `sec 54`
- `sin 54`

The answer is '`cos 54`'

*Solved Exercise Problem: *

`(tan 75) / (cot 15) = ?`

- `tan^2 75`
- `1`
- `1`

The answer is '`1`' As `cot 15` `=tan(90-15)` `=tan 75`.

*Solved Exercise Problem: *

Which of the following equals `sin 36 - cos 36 + sin 54`?

- `sin 54`
- `sin 36`
- `sin 36`

The answer is '`sin 36`'.

*switch to slide-show version*