NCERT Solutions Class 11 Maths
The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 3 Trigonometric Functions Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.
Chapter 3 (Trigonometric Functions)
| Exercise – 3.2 |
Find the values of other five trigonometric functions in Exercises 1 to 5.
1. cos x = -1/2, x lies in third quadrant.
Solution – Since cos x = (-1/2)
We have sec x = 1/cos x = -2
Now sin2 x + cos2 x = 1
i.e., sin2 x = 1 – cos2 x
or sin2 x = 1 – (1/4) = (3/4)
Hence sin x = ±(√3/2)
Since x lies in third quadrant, sin x is negative.
Therefore sin x = (–√3/2)
Which also gives
cosec x = 1/sin x = (–2/√3)
Further, we have
tan x = sin x/cos x
= (–√3/2)/(-1/2)
= √3
and
cot x = 1/tanx = (1/√3)
2. sin x = 3/5, x lies in second quadrant.
Solution – Since sin x = (3/5)
we have cosec x = 1/sin x = (5/3)
Now sin2 x + cos2 x = 1
i.e., cos2 x = 1 – sin2 x
or cos2 x = 1 – (3/5)
= 1 – (9/25)
= (16/25)
Hence cos x = ±(4/5)
Since x lies in second quadrant, cos x is negative.
Therefore
cos x = (–4/5)
which also gives
sec x = 1/cos x= (-5/4)
Further, we have
tan x = sin x/cos x
= (3/5)/(-4/5)
= (-3/4)
and
cot x = 1/tan x = (-4/3)
3. cot x = 3/4, x lies in third quadrant.
Solution – Since cot x = (3/4)
we have tan x = 1 / cot x = (4/3)
Now sec2 x = 1 + tan2 x
= 1 + (16/9)
= (25/9)
Hence sec x = ±(5/3)
Since x lies in third quadrant, sec x will be negative. Therefore
sec x = (-5/3)
which also gives
cos x = (-3/5)
Further, we have
sin x = tan x * cos x
= (4/3) × (-3/5)
= (-4/5)
and
cosec x = 1/sin x
= (-5/4)
4. sec x = 13/5, x lies in fourth quadrant.
Solution – Since sec x = (13/5)
we have cos x = 1/secx = (5/13)
Now sin2 x + cos2 x = 1
i.e., sin2 x = 1 – cos2 x
or
sin2 x = 1 – (5/13)2
= 1 – (25/169)
= 144/169
Hence sin x = ±(12/13)
Since x lies in forth quadrant, sin x is negative. Therefore
sin x = (–12/13)
which also gives
cosec x = 1/sin x = (-13/12)
Further, we have
tan x = sin x/cos x
= (-12/13) / (5/13)
= (-12/5)
and
cot x = 1/tan x
= (-5/12)
5. tan x = -5/12, x lies in second quadrant.
Solution – Since tan x = (-5/12)
we have cot x = 1/tan x = (-12/5)
Now sec2 x = 1 + tan2 x
= 1 + (25/144)
= 169/144
Hence sec x = ±(13/12)
Since x lies in second quadrant, sec x will be negative. Therefore
sec x = (-13/12)
which also gives
cos x = 1/sec x = (-12/13)
Further, we have
sin x = tan x * cos x
= (-5/12) × (-12/13)
= (5/13)
and
cosec x = 1/sin x
= (13/5)
Find the values of the trigonometric functions in Exercises 6 to 10.
6. sin 765°
Solution – We known that the values of sin x repeats after an interval of 2π or 360∘.
So,
sin(765°)
= sin(720° + 45°) { taking nearest multiple of 360 }
= sin(2 × 360° + 45°)
= sin(45°)
= 1/√2
Hence, sin(765°) = 1/√2
7. cosec (–1410°)
Solution – We known that the values of cosec x repeats after an interval of 2π or 360∘.
So, cosec(-1410°)
= – cosec(1410°)
= – cosec(1440° – 30°) { taking nearest multiple of 360 }
= – cosec(4 × 360° – 30°)
= cosec(30°)
= 2
Hence, cosec(–1410°) = 2.
8. tan(19π/3)
Solution – We known that the values of tan x repeats after an interval of π or 180∘.
So,
tan(19π/3)
= tan(18π/3 + π/3) { breaking into nearest integer }
= tan(6π + π/3)
= tan(π/3)
= tan(60°)
= √3
Hence, tan(19π/3) = √3.
9. sin(–11π/3)
Solution – We known that the values of sin x repeats after an interval of 2π or 360∘.
So, sin(-11π/3)
= -sin(11π/3)
= -sin(12π/3 – π/3) { breaking nearest multiple of 2π divisible by 3}
= -sin(4π – π/3)
= -sin(-π/3)
= -[-sin(π/3)]
= sin(π/3)
= sin(60°)
= √3/2
Hence, sin(-11π/3) = √3/2.
10. cot(–15π/4)
Solution – We known that the values of cot x repeats after an interval of π or 180°.
So, cot(-15π/4)
= -cot(15π/4)
= -cot(16π/4 – π/4) { breaking into nearest multiple of π divisible by 4 }
= -cot(4π – π/4)
= -cot(-π/4)
= -[-cot(π/4)]
= cot(45°)
= 1
Hence, cot(-15π/4) = 1.
