Class 10 Trigonometric Ratios of Complementary Angles

### Topic to be covered

☼ Trigonometric Ratios of Complementary Angles.

### 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.
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
Solution:

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
Solution:

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