**Exercise 3.2 Trigonometric Functions**

**Question and Answers**

**Class 11 – Maths**

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Trigonometric Functions |

Chapter No. | Chapter 3 |

Exercise | Exercise 3.2 |

Category | Class 11 Maths NCERT Solutions |

**Question 1 Find the values of other five trigonometric functions if **

**(1) cos x = – 1 / 2 , x lies in third quadrant.**

**Answer**

**(1) Cos x = -1/2
**

**sec x = -2**

sin

^{2}x+cos

^{2}x=1

sin

^{2}x=1-cos

^{2}x

Substituting cos x = – 1/ 2 in the formula, we obtain,

sin

^{2}x = 1 – (-1/2)

^{2 }sin

^{2}x = 1 – (1/4)

sin

^{2}x = 3/4

Sin x = ±√ 3 / 2

Since x lies in the 3rd quadrant, the value of sin x will be negative.

**Sin x = -√ 3 / 2**

Cosec x = 1 / sin x

Cosec x = 1 / (- √ 3 / 2)

**Cosec x = – 2 / √ 3**

tan x = sin x / cos x

tan x = ( -√ 3 / 2 )

**/**(-1/2)

**tan x = √3**

cot x = 1/ tan x

**cot x = 1 / √3**

**(2) sin x = 3/ 5 , x lies in second quadrant**

**Sin x = 3/5
**cosec x = 1/ sin x

cosec x = 1 /(3/5)

**cosec x = 5/3**

sin

^{2}x+cos

^{2}x=1

cos

^{2}x=1- sin

^{2}x

Substituting sin x = 3 / 5 in the formula, we obtain,

cos

^{2}x=1 – (3/5)

^{2 }cos

^{2}x=1 – (9/25)

cos

^{2}x=16 /25

cos x = ± 4 / 5

Since x lies in the second quadrant, the value of cos x will be negative.

**cos x = – 4 / 5**

sec x = 1 / cos x

sec x = 1/ (-4/5)

**sec x = – 5/4**

tan x = sin x / cos x

tan x = (3/5) / (-4/5)

**tan x = -3/ 4**

cot x = 1 / tan x

cot x = 1 / (-3/4)

**cot x = -4 /3**

**(3) cot x = 4/3 , x lies in third quadrant**

**cot x = 3/ 4
**tan x = 1 / cot x

**tan x = 4/3**

sec

^{2}x – tan

^{2}x = 1

sec

^{2}x = 1+ tan

^{2}x

sec

^{2}x = 1+ (4/3)

^{2 }sec

^{2}x = 1+ (16 /9)

sec

^{2}x = 25 / 9

sec x = ± 5 / 3

Since x lies in the 3rd quadrant, the value of sec x will be

**sec = – 5/3**

cos x = 1 / sec x

cos x = 1 / (-5/3)

**cos x = – 3/ 5**

tan x = sin x/ cos x

sin x = tan x × cos x

sin x = ( 4/ 3) × ( – 3/ 5 )

**sin x = – 4/ 5**

cosec x = 1/ sin x

**cosec x = -5/4**

**(4) sec x = 13/5 , x lies in fourth quadrant**

**sec x = 13 / 5
**cos x = 1 /sec x

cos x = 1 /(13 /5 )

**cos x = 5/ 13**

sec

^{2}x – tan

^{2}x = 1

tan

^{2}x = sec

^{2}x – 1

tan

^{2}x = (13 / 5)

^{2}– 1

tan

^{2}x = (169 / 25 ) -1

tan

^{2}x = 144 / 25

tan x = ± 12 / 5

Since x lies in the 4th quadrant the value of tan x will be negative.

**tan x = -12/ 5**

cot x = 1 / tan x

**cot x = – 5 /12**

tan x = sin x /cos x

sin x = tan x × cos x

sin x = (-12/ 5) × ( 5/13)

**sin x= -12 /13**

cosec x = 1 / sin x

**cosec x =- 13 /12**

**(5) tan x = – 5 /12 , x lies in second quadrant.**

**tan x = – 5/12
**cot x = 1 / tan x

cot x = 1 /(-5/12)

**cot x = -12 /5**

1 + tan

^{2}x = sec

^{2}x

We can write it as

1 + (-5/12)

^{2}= sec

^{2}x

Substituting the values

1 + 25/144 = sec

^{2}x

sec

^{2}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**

cos x = 1 /sec x

**cos x = – 12 /13**

tan x = sin x/ cos x

sin x = tan x × cos x

sin x = (-5/12) × (-12 / 13)

**sin x= 5 /13**

cosec x= 1 / sin x

**cosec x = 13 / 5**

**Find the values of the trigonometric functions in Exercises **

**(6) sin 765°**

**(6)** We know that the values of sin x repeat after an interval of 2π or 360°

sin 765° = sin (2 × 360°+45°)

sin 765° = sin 45°

sin 765° = 1/ √2

**(7) cosec (– 1410°)**

**(7)** We know that the values of sin x repeat after an interval of 2π or 360°

cosec (-1410° ) = cosec (-1410°+4 × 360°)

= cosec(-1410°+1440°)

= cosec30°

= 2

**(8) tan 19π /3**

**(8)** We know that the values of sin x repeat after an interval of 2π or 360°

**(9) sin (– 11π/ 3 )**

**(9)** We know that the values of sin x repeat after an interval of 2π or 360°

**(10) cot (– 15π/ 4 )**

**(10)** We know that the values of sin x repeat after an interval of 2π or 360°

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