`☼` Trigonometric Ratios of Complementary Angles.

Recall that two angles are said to be complementary if their sum equals `90°`. In D ABC, right-angled at B, do you see any pair of complementary angles? (See Fig. 8.21)

Since `∠ A + ∠ C = 90°`, they form such a pair. We have:

`tt( (sinA = (BC)/(AC) , cosA = (AB)/(AC), tanA = (BC)/(AB)), (cosecA = (AC)/(BC), secA = (AC)/(AB), cotA = (AB)/(BC)))}` ..............(1)

Now let us write the trigonometric ratios for `∠ C = 90° – ∠ A.`

For convenience, we shall write `90° – A` instead of `90° – ∠ A.`

What would be the side opposite and the side adjacent to the angle `90° – A?`

You will find that AB is the side opposite and BC is the side adjacent to the angle

`90° – A.` Therefore,

`tt((sin(90^0-A) = (AB)/(AC) , cos(90^0-A) = (BC)/(AC) , tan(90^0-A) = (AB)/(BC)) , (cosec(90^0-A) = (AC)/(AB) , sec(90^0-A) = (AC)/(BC) , cot(90^0-A) = (BC)/(AB) )) }` ............(2)

Now, compare the ratios in (1) and (2). Observe that :

`sin(90^0-A) = (AB)/(AC) = cos A` and `cos(90-A) = (BC)/(AC) = sinA`

also `tan(90^0-A) = (AB)/(BC) = cotA , cot(90^0-A) = (BC)/(AB) = tanA`

`sec(90^0-A) = (AC)/(BC) = cosecA , cosec(90^0-A) = (AC)/(AB) = secA`

so `sin(90^0-A) = cosA, \ \ \ \ \ \ \ \ cos(90^0-A) = sinA`

`tan(90^0-A) = cotA ,\ \ \ \ \ \ \ \ \ \ cot(90^0-A) = tanA`

`sec(90^0-A) = cosecA , \ \ \ \ \ \ \ \ \ cosec(90^0-A) = secA`

for all values of angle A lying between `0°` and `90°`. Check whether this holds for `A = 0°` or `A = 90°`.

`"Note :"` tan `0° = 0 = cot 90°, sec 0° = 1 = cosec 90°` and `sec 90°, cosec 0°, tan 90°` and

`cot 0°` are not defined.

Since `∠ A + ∠ C = 90°`, they form such a pair. We have:

`tt( (sinA = (BC)/(AC) , cosA = (AB)/(AC), tanA = (BC)/(AB)), (cosecA = (AC)/(BC), secA = (AC)/(AB), cotA = (AB)/(BC)))}` ..............(1)

Now let us write the trigonometric ratios for `∠ C = 90° – ∠ A.`

For convenience, we shall write `90° – A` instead of `90° – ∠ A.`

What would be the side opposite and the side adjacent to the angle `90° – A?`

You will find that AB is the side opposite and BC is the side adjacent to the angle

`90° – A.` Therefore,

`tt((sin(90^0-A) = (AB)/(AC) , cos(90^0-A) = (BC)/(AC) , tan(90^0-A) = (AB)/(BC)) , (cosec(90^0-A) = (AC)/(AB) , sec(90^0-A) = (AC)/(BC) , cot(90^0-A) = (BC)/(AB) )) }` ............(2)

Now, compare the ratios in (1) and (2). Observe that :

`sin(90^0-A) = (AB)/(AC) = cos A` and `cos(90-A) = (BC)/(AC) = sinA`

also `tan(90^0-A) = (AB)/(BC) = cotA , cot(90^0-A) = (BC)/(AB) = tanA`

`sec(90^0-A) = (AC)/(BC) = cosecA , cosec(90^0-A) = (AC)/(AB) = secA`

so `sin(90^0-A) = cosA, \ \ \ \ \ \ \ \ cos(90^0-A) = sinA`

`tan(90^0-A) = cotA ,\ \ \ \ \ \ \ \ \ \ cot(90^0-A) = tanA`

`sec(90^0-A) = cosecA , \ \ \ \ \ \ \ \ \ cosec(90^0-A) = secA`

for all values of angle A lying between `0°` and `90°`. Check whether this holds for `A = 0°` or `A = 90°`.

`"Note :"` tan `0° = 0 = cot 90°, sec 0° = 1 = cosec 90°` and `sec 90°, cosec 0°, tan 90°` and

`cot 0°` are not defined.

Q 3210501410

If `sin 3A = cos (A – 26°)`, where `3A` is an acute angle, find the value of A

Class 10 Chapter 8 Example 10

Class 10 Chapter 8 Example 10

We are given that `sin 3A = cos (A – 26°)`. ...............(1)

Since `sin 3A = cos (90° – 3A)`, we can write (1) as

`cos (90° – 3A) = cos (A – 26°)`

Since `90° – 3A` and `A – 26°` are both acute angles, therefore,

`90° – 3A = A – 26°`

which gives `A = 29°`

Q 3220501411

Express `cot 85° + cos 75°` in terms of trigonometric ratios of angles between `0°` and `45°`.

Class 10 Chapter 8 Example 11

Class 10 Chapter 8 Example 11

`cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°)`

`= tan 5° + sin 15°`