Chapter 3: Trigonometric Functions Exercise 3.2

It is given that

sin x = 3/5

We can write it as

We know that

sin² x + cos² x = 1

We can write it as

cos² x = 1 – sin² x

It is given that

cot x = 3/4

We can write it as

We know that

1 + tan² x = sec² x

We can write it as

1 + (4/3)² = sec² x

Substituting the values

1 + 16/9 = sec² x

cos² x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant, so the value of sec x will be negative

sec x = – 5/3

We can write it as

It is given that

sec x = 13/5

We can write it as

We know that

sin² x + cos² x = 1

We can write it as

sin² x = 1 – cos² x

Substituting the values

sin² x = 1 – (5/13)²

sin² x = 1 – 25/169 = 144/169

sin² x = ± 12/13

Here x lies in the fourth quadrant, so the value of sin x will be negative

sin x = – 12/13

We can write it as

It is given that

tan x = – 5/12

We can write it as

We know that

1 + tan² x = sec² x

We can write it as

1 + (-5/12)² = sec² x

Substituting the values

1 + 25/144 = sec² x

sec² x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant, so the value of sec x will be negative

sec x = – 13/12

We can write it as

Find the values of the trigonometric functions in Exercises 6 to 10.

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

= sin 45^o

= 1/ √ 2

We know that values of cosec x repeat after an interval of 2π or 360°

So we get

By further calculation

= cosec 30^o = 2

8.

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation

We get

= tan 60^o

= √3

9.

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

10.

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation