# Class 11 NCERT Solutions – Chapter 3 Trigonometric Function – Exercise 3.2

• Last Updated : 15 Dec, 2020

### Question. Find the values of the other five trigonometric functions in Exercises 1 to 5.

1. cos x = –1/2, x lies in the 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

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 the 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 the 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) x (-3/5)

= (-4/5)

and cosec x = 1/sin x

= (-5/4)

4. sec x = 13/5, x lies in the 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 the 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) x (-12/13)

= (5/13)

and cosec x = 1/sin x = (13/5)

### Question. 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 nearset 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 nearset multiple of π  divisible by 4 }

= -cot(4π – π/4)

= -cot(-π/4)

= -[-cot(π/4)]

= cot(45°)

= 1

Hence, cot(-15π/4) = 1.

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