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Chapter 3 Trigonometric Functions Ex 3.2 Solutions

Question - 1 : - Find the values of other five trigonometric functions if cos x = -1/2, x lies in third quadrant.

Answer - 1 : -


Question - 2 : - Find the values of other five trigonometric functions if sin x = 3/5, x lies in second quadrant.

Answer - 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

Question - 3 : - Find the values of other five trigonometric functions if cot x = 3/4, x lies in third quadrant.

Answer - 3 : -

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 thevalues

1 + 16/9 = sec2 x

cos2 x= 25/9

sec x = ± 5/3

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

sec x = – 5/3

We can write it as

Question - 4 : - Find the values of other five trigonometric functions if sec x = 13/5, x lies in fourth quadrant.

Answer - 4 : -

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 thevalues

sin2 x= 1 – (5/13)2

sin2 x= 1 – 25/169 = 144/169

sin2 x= ± 12/13

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

sin x = – 12/13

We can write it as

Question - 5 : - Find the values of other five trigonometric functions if tan x = -5/12, x lies in second quadrant.

Answer - 5 : -

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 thevalues

1 + 25/144 = sec2 x

sec2 x= 169/144

sec x = ± 13/12

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

sec x = – 13/12

We can write it as

Question - 6 : - Find the value of the trigonometric function sin 765°

Answer - 6 : -

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

So we get

By further calculation

= sin 45o

= 1/ √ 2

Question - 7 : - Find the value of the trigonometric function cosec (–1410°)

Answer - 7 : -

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

So we get

By further calculation

= cosec 30o =2

Question - 8 : - Find the value of the trigonometric function 

Answer - 8 : -

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

So we get

By further calculation

We get

= tan 60o

= √3

Question - 9 : - Find the value of the trigonometric function 

Answer - 9 : -

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

So we get

By further calculation

Question - 10 : - Find the value of the trigonometric function

Answer - 10 : -

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

So we get

By further calculation

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