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.